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CBSE Questions for Class 11 Engineering Physics Motion In A Straight Line Quiz 7 - MCQExams.com
CBSE
Class 11 Engineering Physics
Motion In A Straight Line
Quiz 7
A car accelerates from rest at constant rate of $$2 m/s^2$$ for some time. Then its retards at a constant rate of $$4 m/s^2$$ and comes to rest. if it remains in motion for $$3$$ seconds, then the maximum speed attained by the car is:-
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$$2 m/s$$
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$$3 m/s$$
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$$4 m/s$$
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$$6 m/s$$
Explanation
Let $$v, t_1,t_2$$ be the maximum velocity attained, time for which car accelerated and the time for which car retarded respectively
$$\Rightarrow t_1=(v-u)/a=(v-0)/2=v/2$$
$$\Rightarrow t_2=(v-u)/a=(0-v)/(-4)=v/4$$
Now it is given that,
$$\Rightarrow t_1+t_2=3$$
$$\Rightarrow v/2+v/4=3$$
$$\Rightarrow v=4m/s$$
Hence correct answer is option $$C $$
A particle is projected vertically upwards and it attains maximum height $$H$$. If the ratio of times to attain height $$h(h < H)$$ is $$1/3$$, then $$h$$ equals
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2/3 H
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3H
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5/9. H
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3/. H
Explanation
Total time taken to reach height H$$=t=\sqrt{\dfrac{2H}{g}}$$
Initial velocity$$=u=\sqrt{2gH}$$
Time to reach height h$$=t_1=t/3=\dfrac{1}{3} \sqrt{\dfrac{2H}{g}}$$
Now by second equation of motion
$$\Rightarrow h=ut_1+0.5\times(-g)(t_1^2)$$
$$\Rightarrow h=\sqrt{2gH}\times $$
$$\dfrac{1}{3}\times \sqrt{\dfrac{2H}{g}}$$
$$-0.5\times g \times {(\dfrac{1}{3} \sqrt{\dfrac{2H}{g}})}^2$$
$$\Rightarrow h=\dfrac{2H}{3}-\dfrac{H}{9}=\dfrac{5H}{9}$$
$$\Rightarrow \dfrac{h}{H}=\dfrac{5}{9}$$
Hence none of the options are correct.
Two particles P and Q start from rest and move for equal time on a straight line. Particle P has an acceleration of$$X m/s^2$$ for the first half of the total time and $$2x m/s^2$$ for the second half. Particle Q has an acceleration of $$2X m/s^2$$ for the first half of the total time and $$X m/s^2$$ for the second half. Which particle has covered larger distance?
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Both have covered the same distance
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P has covered the larger distance
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Q has covered the larger distance
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data insufficient
Explanation
assuming total time is $$2t$$
for $$P$$, at half time
$$v=X.t, S_1=\frac{1}{2}Xt^2$$
at full time $$S_2=Xt.t+\frac{1}{2}2Xt^2=2Xt^2$$
total distance covered by $$P$$ is $$ S_1+S_2=2.5Xt^2$$
for particle $$Q$$, at half time
$$v=2Xt,S_1=\frac{1}{2}2X t^2=Xt^2$$
at full time $$S_2=2Xt.t+\frac{1}{2}Xt^2=2.5Xt^2$$
total distance covered by $$Q$$ is $$ S_1+S_2=3.5Xt^2$$
so distance covered by $$Q $$ is more then $$P$$.
correct answer is C.
From the top of a tower, a stone is thrown up. It reaches the ground in $$5$$ $$s$$. A second stone is thrown down with the same speed and reaches the ground in $$1$$ $$s$$. A third stone is released from rest and reaches the ground in
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3
$$s$$
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2
$$s$$
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$$\sqrt 5$$
$$s$$
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2.5
$$s$$
Explanation
let height of tower is $$H$$, acceleration due to gravity is $$a$$,
case 1) initial velocity is $$-u$$ (assuming downward direction as positive.
then $$H=-ut+\frac{1}{2}at^2 =>H=-5u+\frac{25}{2}a$$
for case 2) initial velocity is $$u$$
then $$H=ut+\frac{1}{2}at^2=>H=u+\frac{1}{2}a$$
multiplying 2nd with 5 and
adding both equation we get
$$6H=15a=>H=2.5a$$
now for 3rd case $$u=0$$, we get $$2.5a=\frac{1}{2}at^2=>t=\sqrt5 sec.$$
A flowerpot falls from a windowsill $$25.0 m$$ above the sidewalk. How fast is the flowerpot moving when it strikes the ground? (in m/s)
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$$22$$
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$$23$$
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$$26$$
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$$25$$
Explanation
Initial speed of flowerpot $$u=0$$
Height $$h = 25.0 \ m$$
Applying $$3_{rd}$$ equation of motion,
$$v^{2}=u^{2}+2 gh$$
$$v^{2}=0^{2}+ 2 \times 9.8 \times25$$
$$v=22 \ m/s$$
What is the total height reached by the stone? (in m)
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$$36$$
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$$35$$
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$$34.44$$
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$$39$$
Explanation
Maximum height in verticle motion is given by $$H=\frac{u^2}{2g}$$
Also by using $$v^2-u^2=2aS$$
Given v=15, S=$$\dfrac{2H}{3}$$, using $$a=g=-9.81$$
We get $$u^2=225+13.08H$$
Solving both equations eleminating $$u^2$$
We get
$$2gH=225+13.08H$$
$$H=34.403$$
So closest option is optionC.
Maximum height. (in m)
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$$20$$
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$$40$$
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$$60$$
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$$80$$
Explanation
Let the maximum height reached be $$H$$.
At the maximum height, final velocity of ball is zero i.e. $$v = 0$$
Initial velocity $$u = +20 \ m/s$$ (Considering upward direction to be positive)
Acceleration due to gravity $$g = -9.8 \ m/s^2$$
Using $$v^2 - u^2 =2gH$$
Or $$0-20^2 = 2(-9.8)H$$
$$\implies\ H \approx 20 \ m$$
An astronaut jumps from an airplane. After he had fallen $$40 m$$, then his parachute opens. Now he falls with a retardation of $$2 m/s^2$$ and reaches the earth with a velocity of $$3.0 m/s$$. What was the height of the aeroplane? (in m)
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$$185$$
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$$190$$
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$$234$$
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$$200$$
Explanation
Velocity after falling through 40m=$$\sqrt{2gh}=\sqrt{2\times 9. 81 \times 40}=28m/s$$
Distance travelled after opening the parachute=$$(28^2-3^2)/2\times 2=194$$
So, height of the aeroplane$$=194+40=234m$$
Hence none of the above options are correct.
A car starts from rest and accelerates uniformly to a speed of $$ 180 \,km \,h^{-1} $$ in $$10$$ sec. The distance covered by the car in the time interval is :
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$$200\, m$$
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$$300\, m$$
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$$500 \,m$$
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$$250\, m$$
Explanation
$$u = 0$$,
$$v = 180$$
$$180km/h = 50 m/s$$
Time taken, $$t = 10 s$$
$$ a = \dfrac {v-u}{t} = \dfrac {60}{10} = 6 ms^{-2} $$
Distance covered
$$ s = ut + \dfrac {1}{2} at^2 $$
$$ = 0 + \dfrac {1}{2} \times 5 \times ( 10 )^2 $$
$$ = \dfrac {500}{2} \Rightarrow = 250 m $$
Two stones of mass $$m_1$$ and $$m_2$$ (such that $$m_1$$ and $$m_2$$ ) are dropped $$\triangle$$ t time apart from the same height towards the ground. At a later time t the difference in their speed is $$\triangle $$ V and their mutual separation is $$\triangle$$ S . While both stones are in flight
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$$\triangle$$ V decreases with time and $$\triangle$$ S increases with time.
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$$\triangle$$ V and $$\triangle$$ S increases with time.
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$$\triangle$$ V remains constant time and $$\triangle$$ S decreases with time.
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$$\triangle$$ V remains constant time and $$\triangle$$ S increases with time.
Explanation
$$\Delta V=g\Delta t$$
Since $$\Delta t$$ is unchanged, $$\Delta V$$ remains same.
After some time t,
$$\Delta S=\frac{gt^2}{2}-\frac{g(t-\Delta t)^2}{2}=\frac{1}{2}(2gt\Delta t+(\Delta t)^2)$$
$$\Delta S \propto t$$ so $$\Delta S$$ increases with time.
A particle moves along the x-axis with a position given by the equation $$x = 5 + 3t$$, where $$x$$ is in meters, and $$t$$ is in seconds. The positive direction is east. Which of the following statements about the particle is false?
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The particle is east of the origin at $$t = 0$$
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The particle is at rest at $$t = 0$$
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The particle's velocity is constant
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The particle's acceleration is constant
Explanation
Given position vector is: $$x = 5 + 3t$$
Hence, at $$t = 0, \quad x = 5\ m$$ i.e., particle is east of origin. Option A is correct.
Comparing with second equation of motion,
$$s = ut + \cfrac{1}{2}at^2$$
$$u = 3\ m/s$$
Option B is false
$$a = 0$$ Thus, option D is true.
Hence, particle's velocity is constant. O
ption C is true.
A car moving with a velocity of $$20{ ms }^{ -1 }$$ is stopped in a distance of $$40 m$$. If the same car is travelling at double the velocity, the distance travelled by it for same retardation is
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$$640 m$$
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$$320 m$$
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$$1280 m$$
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$$160 m$$
Explanation
Initial speed of the car is $$u$$.
Final speed of the car $$v=0$$
Let the retardation of the car be $$a$$.
Using $$v^2 - u^2 = 2aS$$
Or $$0- u^2 = 2aS$$
$$\implies$$ $$S\propto u^2$$ (for same $$a$$)
Given : $$S = 40 m$$ $$u' = 2u$$
$$\therefore$$ $$\dfrac{S'}{S} = \dfrac{(u')^2}{u^2} =2^2$$
Or
$$\dfrac{S'}{40} = 4$$
$$\implies$$ $$S' = 160 m$$
A body travelling along a straight line one-third of the total distance with a velocity $$4m/s$$. The remaining part of the distance was covered with a velocity $$2m/s$$ for the first half of the remaining journey and with a velocity $$6m/s$$ for the another half of the remaining journey. The average velocity is
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$$5m/s$$
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$$4m/s$$
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$$4.5m/s$$
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$$3.5m/s$$
Explanation
Let total distance d
Let time covered for first one third distance be$$t_1$$
Then $$t_1=\large\frac{d/3}{4}=\large\frac{d}{12}$$
$$t_2$$ be time for next two journeys
$$d-\frac{d}{3}=\frac{2d}{3}=2t_2+6 t_2=8t_2$$
$$t_2=\large\frac{d}{12}$$
$$Average\;velocity=\large\frac{Total\;distane}{Total\;time}=\frac{d}{t_1+2t_2}=\large\frac{d}{\Large\frac{d}{12}+\frac{2d}{12}}=4m/s$$
A juggler tosses a ball up in the air with initial speed $$u$$. At the instant it reaches its maximum height $$H$$, he tosses up a second ball with the same initial speed. The two balls will collide at a height
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$$\dfrac { H }{ 4 } $$
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$$\dfrac { H }{ 2 } $$
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$$\dfrac { 3H }{ 4 } $$
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$$\sqrt { \dfrac { 3 }{ 4 } } H$$
Explanation
Step 2: Finding the equation for height.
Step 3: Finding the value of height.
A particle moving in one dimension with a constant acceleration of $$2 {m}/{{s}^{2}}$$ is observed to cover a distance of $$5 m$$ during a particular interval of $$1 s$$. The distance covered by the particle in the next $$1 s$$ interval is in metre
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$$5$$
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$$6$$
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$$7$$
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$$10$$
Explanation
The distance cover 5 m in time interval 1 s means its speed is $$5 m/s$$.
Using, $$v=at$$, $$t=v/a=5/2 s$$
For next interval, $$t_1=t+1=5/2+1=7/2 s$$
For after time $$t_1$$, the speed will be $$v_1=at_1=2(7/2)=7 m/s$$
Thus, distance covered by next 1 s interval will be $$7 m$$.
In a car race, car $$A$$ takes a time $$t$$ less than car $$B$$ at the finish and passes the finishing point with speed $$v$$ more than that of the car $$B$$. Assuming that both the cars starts from rest and travel with constant accelerations $${ a }_{ 1 }$$ and $${ a }_{ 2 }$$ respectively. So, the value of $$v$$ will be :
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$$\left( \sqrt { { { a }_{ 1 } }/{ { a }_{ 2 } } } \right) t$$
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$$\left( \sqrt { { { a }_{ 2 } }/{ { a }_{ 1 } } } \right) t$$
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$$\left( { a }_{ 1 }\sqrt { { a }_{ 2 } } \right) t$$
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$$\left( \sqrt { { a }_{ 1 }{ a }_{ 2 } } \right) t$$
Explanation
Consider that $$A$$ takes $${ t }_{ 1 }$$ second, then according to the given problem, $$B$$ will take $$\left( { t }_{ 1 }+t \right) $$ seconds. Further let $${ v }_{ 1 }$$ be the velocity of $$B$$ at finishing point, then velocity of $$A$$ will be $$\left( { v }_{ 1 }+v \right)$$. Writing equations of motion for $$A$$ and $$B$$
$${ v }_{ 1 }+v={ a }_{ 1 }{ t }_{ 1 }$$ ....(i)
$${ v }_{ 1 }={ a }_{ 2 }\left( { t }_{ 2 }+t \right) $$ ....(ii)
From equations (i) and (ii), we get
$$v=\left( { a }_{ 1 }-{ a }_{ 2 } \right) { t }_{ 1 }-{ a }_{ 2 }t$$ .....(iii)
Total distance travelled by both the cars is equal
$${ S }_{ A }={ S }_{ B }$$
$$\Rightarrow \dfrac { 1 }{ 2 } { a }_{ 1 }{ t }_{ 1 }^{ 2 }=\dfrac { 1 }{ 2 } { a }_{ 2 }{ \left( { t }_{ 1 }+t \right) }^{ 2 }\Rightarrow { t }_{ 1 }=\dfrac { \sqrt { { a }_{ 2 } } t }{ \sqrt { { a }_{ 1 } } -\sqrt { { a }_{ 2 } } }$$
Substituting this value of $${ t }_{ 1 }$$ in equation (iii), we get
$$v=\left( \sqrt { { a }_{ 1 }{ a }_{ 2 } } \right) t$$
A car is moving along a straight road with a uniform acceleration. It passes through two points $$P$$ and $$Q$$ separated by a distance, with velocity $$30\ km/hr$$ and $$40\ km/hr$$ respectively. The velocity of the car midway between $$P$$ and $$Q$$ is
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$$35.35\ km/hr$$
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$$23.3\ km/hr$$
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$$283\sqrt {2} km/hr$$
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none of these.
Explanation
using equations of motion $$v^2-u^2=2aS$$
$$v=40,u=30,a=constant,S=?$$
we get $$S=\dfrac{40^2-30^2}{2a}=>S=\dfrac{350}{a}$$
now given $$S_{middle}=\dfrac{1}{2}\times \dfrac{350}{a},a=constant,u=30,v=?$$
$$v_{middle}^2=2aS_{middle}+u^2=>v_{middle}^2=2a\times \dfrac{350}{2a}+30^2=>v_{middle}=\sqrt{350+900}$$
$$v_{middle}=35.35km/hr$$
A body $$X$$ is projected upwards with a velocity of $$98\ ms^{-1}$$, after $$4s$$, a second body $$Y$$ is also projected upwards with the same $$Y$$ is also projected upwards with the same initial velocity. Two bodies will meet after
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$$8\ s$$
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$$10\ s$$
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$$12\ s$$
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$$14\ s$$
Explanation
Let $$t$$ second be the time of flight of the first body after meeting, then $$(t - 4)$$ second will be the time of flight of the second body.
Since, $$h_{1} = h_{2}$$
$$\therefore 98t - \dfrac {1}{2}gt^{2} = 98 (t - 4)g (t - 4)^{2}$$
On solving, $$t = 12s$$.
A man of mass $$60$$kg and a boy of mass $$30$$kg are standing together on frictionless ice surface. If they push each other apart man moves away with a speed of $$0.4$$m/s relative to ice. After $$5$$sec they will be away from each other at a distance of.
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$$9.0$$m
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$$3.0$$m
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$$6.0$$m
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$$30$$,
Explanation
The man and the boy move in opposite directions.
Momentum of man$$=$$ momentum of boy
$$60\times 0.4=30\times v$$
or velocity of the boy $$v=0.8ms^{-1}$$
$$\therefore$$ Relative velocity $$=0.4+0.8=1.2ms^{-1}$$
$$\therefore$$ Distance between them in $$5$$sec $$ =1.2\times 5=6.0$$m.
If the distance travel by a uniformly accelerated particle in $$pth, qt$$ and $$rth$$ second are $$a, b$$ and $$c$$ respectively. Then
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$$(q - r) a + (r - p) b + (p - q) c = 1$$
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$$(q - r) a + (r - p) b + (p - q) c = -1$$
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$$(q - r) a + (r - p) b + (p - q) c = 0$$
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$$(q + r) a + (r + p) b + (p + q) c = 0$$
Explanation
Let $$u$$ be the initial velocity of the particle and $$f$$ be acceleration of the particle. Then
$$a = u + \dfrac {1}{2} f(2p - 1) ..... (i)$$
$$b = u + \dfrac {1}{2}f(2q - 1) .... (ii)$$
$$c = u + \dfrac {1}{2} f(2r - 1) ..... (iii)$$
Now, multiplying Eq. (i) by $$(q - r)$$, Eq. (ii) by $$(r - p)$$ and Eq. (iii) by $$(p - q)$$ and adding, we get $$a(q - r) + b(r - p) + c(p - q) = a(0) + \dfrac {1}{2} f(0)$$
$$\therefore a(q - r) + b(r - p) + c(p - q) = 0$$.
If a coin is dropped in a lift it takes $$t_{1}$$ time to reach the floor and takes $$t_{2}$$ time when lift is moving up with constant acceleration, when which one of the following relation is correct?
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$$t_{1} = t_{2}$$
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$$t_{1} > t_{2}$$
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$$t_{2} > t_{1}$$
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$$t_{1} > > t_{2}$$
Explanation
Time $$t_{1}$$ for stationary lift $$= \sqrt {\dfrac {2h}{g}}$$
When lift is moving up with constant acceleration, then
$$t_{2} = \sqrt {\dfrac {2h}{g + a}}$$
$$\therefore t_{1} > t_{2}$$.
A body starts from rest acquires a velocity $$v$$ in time $$T$$. The work done on the body in time$$t$$ will be proportional to
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$$\left( \cfrac { { V }^{ 2 } }{ { T }^{ 2 } } \right) { t }^{ 2 }$$
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$$\left( \cfrac { { V }^{ 2 } }{ { T }^{ 2 } } \right) { t }$$
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$$\left( \cfrac { { V }^{ 2 } }{ { T }^{ } } \right) { t }^{ 2 }$$
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$$\left( \cfrac { { V }^{ } }{ { T }^{ } } \right) { t }^{ 2 }$$
Explanation
The acceleration of the body $$a=\cfrac { v }{ T } $$
Now, work done $$=\cfrac { 1 }{ 2 } m{ v }^{ 2 }\propto { v }^{ 2 }$$
Here $$v=u+at$$
so, work done $${ v }^{ 2 }\propto { a }^{ 2 }{ t }^{ 2 }$$
or $$W\propto \left( \cfrac { { v }^{ 2 } }{ { T }^{ 2 } } \right) { t }^{ 2 }\quad $$
A man weighing 80 kg is standing on a trolley weighing 320 kg. The trolley is resting on frictionless horizontal rails. If the man starts walking on the trolley along the rails at speed 1 m/s. Then after 4 s, his displacement relative to ground will be :
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4.5 m
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5 m
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8 m
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3.2 m
Explanation
In $$1s$$, man has walked, $$4m/s\times1s = 4m$$, with respect to the trolley.
As position of the CoM stays same, let the distance moved by trolley be $$x$$ in direction opposite to that of man.
So net diplacement of man is $$(4-x)m$$
Thus, $$M_{man}\times(4-x) = M_{trolley}x$$
$$\implies 80(4-x) = 320x\implies x=320/400=0.8m$$
So net displacement of man in $$1s$$ is, $$4-0.8 = 3.2m$$
Option D is correct.
Initial speed of an alpha particle side a tube a length $$4\ m$$ is $$1\ km/s$$, if it is accelerated in the tube and comes out with a speed of $$9\ km/s$$, then the time for which the particle remains inside the tube is
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$$8\times 10^{-3}s$$
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$$8\times 10^{-4}s$$
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$$80\times 10^{-3}s$$
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$$800\times 10^{-3}s$$
Explanation
Initial speed, $$u = 1\ km/s = 1000\ m/s$$
Final speed, $$v = 9\ km/s = 9000\ m/s$$
By using the relation
$$v^{2} = u^{2} + 2as$$
$$(9000)^{2} = (1000)^{2} + 2\times a \times 4$$
$$a = 10^{7} m/s^{2}$$
$$\therefore$$ the time for which the particle remains in the tube
$$v = u + at$$
or $$t = \dfrac {v - u}{a} = \dfrac {9000 - 1000}{10^{7}}$$
$$= 8\times 10^{-4}s$$.
The time required to stop a car of mass $$800\ kg$$ moving at a speed of $$20\ ms^{-1}$$ over a distance of $$25\ m$$ is
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$$2\ s$$
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$$2.5\ s$$
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$$4\ s$$
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$$4.5\ s$$
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$$1\ s$$
Explanation
Given, $$u = 0\ ms^{-1}, v = 20\ ms^{-1}$$
Mass of car $$= 800\ kg, s = 25\ m$$
We know that, distance covered,
$$s = \left (\dfrac {u + v}{2}\right ) t$$
$$\Rightarrow 25 = \left (\dfrac {0 + 20}{2}\right )t$$
$$= \dfrac {25}{10} = 2.5 s$$.
A body freely falling from the rest has a velocity v after his falls through a height h. The distance, it has to fall down further for its velocity to become double, is :
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4 h
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3h
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h
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16h
Explanation
$$\textbf{Hint:}$$ Apply third equation of motion.
$$\textbf{Step 1-Use third equation of motion }$$
If a body falls freely so its initial velocity is $$u=0$$
Acceleration is $$a=-g$$ and displacement $$s=-h$$
After travelling distance h velocity becomes v.
So $$v^2=u^2+2as$$
$$\Rightarrow v^2=0+2(-g)(-h)=2gh$$.......(1)
$$\textbf{Step 2-Calculate distance when velocity becomes double}$$
After travelling distance $$h_1$$,velocity becomes double i.e.$$2v$$.
Again apply third equation of motion
$$(2v)^2=v^2+2(-g)(-h_1)$$
$$\Rightarrow 4v^2=v^2+2gh_1$$.......(2)
$$\textbf{Step 3-Solve above equations to obtain answer}$$
Put value from equation (1)
$$4(2gh)=(2gh)+2gh_1$$
$$\Rightarrow 2gh_1=6gh$$
$$\Rightarrow h_1=3h$$
Hence the correct answer is option (B)
The driver of an express train suddenly sees the red light signal 50 m ahead and applies the brakes .If the average deceleration during braking is $$10.0 ms^{-2}$$ and the reaction time of the driver is 0.75 sec , the minimum speed at which the train should be moving so as not to cross the red signal is
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27 km/hr
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144 km/hr
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72 km/hr
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83 km/hr
Explanation
Let,
Displacement of Train in 0.75 sec will be
$$S_1 = ut = 0.75 \ u \ m$$
where, u - initial velocity of train.
$$v^2 = u^2 + 2aS$$
$$v = 0 ms^{-1}, S = 50- S_1$$
$$u^2 - 2 \times 10 \times(50 - 0.75u) = 0 $$
solving for u we get
$$u = 144 \ km/hr$$
Three particles $$A, B$$ and $$C$$ are thrown from the top of a tower with the same speed. $$A$$ is thrown straight up, $$B$$ is thrown straight down and $$C$$ is thrown horizontally. They hit the ground with speed $$v_{A}, v_{B}$$ and $$v_{C}$$ respectively.
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$$v_{A} = v_{B} = v_{C}$$
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$$v_{A} > v_{B} > v_{C}$$
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$$v_{A} = v_{B} > v_{C}$$
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$$v_{A} > v_{B} = v_{C}$$
Explanation
For both $$A$$ and $$B, u_{vertical}$$ is the same, because when a ball is projected upwards with an initial velocity, it will come back with the same velocity. At $$Q$$ both have the same initial velocity. For $$C$$, its vertical velocity is zero but it has a horizontal velocity $$u_{C} = u_{A} = u_{B}$$. It falls through the height freely. But $$v_{A}$$ and $$v_{B}$$ will be greater than the final velocity $$v_{C}$$, because $$v_{A} = u_{A} + gt; v_{B} = u_{B} + gt$$. But for $$C, u_{C} + (gt)$$ are not in the same direction. It is less, but $$v_{A} = v_{B} > v_{C}$$.
The deceleration of a car traveling on a straight highway is a function of its instantaneous velocity $$v$$ given by $$w=a\sqrt { v } $$, where $$a$$ is a constant. If the initial velocity of the car is $$v_o$$, the distance the car will travel and the time it takes before it stops are
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$$\cfrac { 2 }{ 3 } m v_o^{3/2},\cfrac { 1 }{ 2 } v_o^{3/2} s$$
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$$\cfrac { 3 }{ 2a } m v_o^{1/2},\cfrac { 1 }{ 2a } v_o^{1/2}s$$
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$$\cfrac { 3a }{ 2 } m v_o^{1/2},\cfrac { a }{ 2 } v_o^{3/2}s\quad $$
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$$\cfrac { 2 }{ 3a } m v_o^{3/2},\cfrac { 2 }{ a } v_o^{1/2} s\quad $$
Explanation
Deceleration of the car is given as $$w = a\sqrt{v}$$
Using $$w = \dfrac{-vdv}{ds}$$
$$\therefore$$ $$\int^o_{v_o} v^{1/2}dv =- a\int^S_o ds$$
We get $$\dfrac{2}{3} v_{o}^{3/2} = aS$$
$$\implies \ S = \dfrac{2}{3a}v_o^{3/2}$$
Using $$w =- \dfrac{dv}{dt}$$
$$\therefore$$ $$-\dfrac{dv}{dt} = a\sqrt{v}$$
Integrating $$\int^o_{v_o} v^{-1/2} dv = -a\int_o^t dt$$
we get, $$2\sqrt{v_o} = at$$
$$\implies \ t = \dfrac{2\sqrt{v_o}}{a}$$
A ball thrown upward from the top of a tower with speed $$v$$ reaches the ground in $$t_{1}$$ second. If this ball is thrown downward from the top of the same tower with speed $$v$$ it reaches the ground in $$t_{2}$$ second. In what time the ball shall reach the ground if it is allowed to fall freely under gravity from the top of the tower?
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0%
$$\dfrac {t_{1} + t_{2}}{2}$$
0%
$$\dfrac {t_{1} - t_{2}}{2}$$
0%
$$\sqrt {t_{1}t_{2}}$$
0%
$$t_{1} + t_{2}$$
Explanation
$$h = -vt_{1} + \dfrac {1}{2} gt_{1}^{2}, h = vt_{2} + \dfrac {1}{2} gt_{2}^{2}$$
Now, $$\dfrac {h}{t_{1}} + \dfrac {h}{t_{2}} = \dfrac {1}{2}g(t_{1} + t_{2})$$ or $$h = \dfrac {1}{2} gt_{1}t_{2}$$
Also, $$h = \dfrac {1}{2}gt^{2} \therefore t^{2} = t_{1}t_{2}$$or $$t = \sqrt {t_{1}t_{2}}$$.
A $$20\ kg$$ bullet pierces through a plate of mass $$M_{1} = 1\ kg$$ and then comes to rest inside a second plate of mass $$M_{2}= 2.98\ kg$$ as shown in the figure. It is found that the two plates initially at rest and now move with equal velocities. Find the percentage loss in the initial velocity of the bullet when it is between $$M_{1}$$ and $$M_{2}$$. (Neglect any loss of material of the plates due to the action of bullet).
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0%
$$6$$%
0%
$$25$$%
0%
$$50$$%
0%
$$72.5$$%
Explanation
Let initial velocity of bullet$$={ V }_{ 1 }㎧$$
Velocity with which each plate moves$$={ V }_{ 2 }㎧$$
According to the law of conservation of momentum the initial momentum of the bullet is equal to the sum of the final momentum of the second plate including the bullet
$$\therefore m{ V }_{ 1 }={ M }_{ 1 }{ V }_{ 2 }+\left( { M }_{ 2 }+m \right) { V }_{ 2 }$$
$$20{ V }_{ 1 }=1\times { V }_{ 2 }+\left( 2.98+20 \right) { V }_{ 2 }$$
$$20{ V }_{ 1 }={ V }_{ 2 }+22.98{ V }_{ 2 }$$
$${ V }_{ 1 }=\cfrac { 23.98 }{ 20 } { V }_{ 2 }$$
$${ V }_{ 1 }=1.199{ V }_{ 2 }\rightarrow 1$$
Let the velocity of bullet when it comes out of first plate$$={ V }_{ 3 }$$
The momentum of the bullet the first and second, a plate is equal to the sum of the momentum of the second plate and the bullet.
$$20{ V }_{ 3 }=\left( 20+2.980{ V }_{ 2 } \right) $$
$$20{ V }_{ 3 }=22.98{ V }_{ 2 }$$
$$\therefore { V }_{ 3 }=1.149{ V }_{ 2 }\rightarrow 2$$
Loss Percentage in the initial velocity of the bullet when it is moving between $${ m }_{ 1 }\& { m }_{ 2 }$$ is expressed as the following
Loss %$$=\cfrac { { V }_{ 1 }-{ V }_{ 3 } }{ { V }_{ 1 } } \times 100$$
Loss %$$=\cfrac { 1.199{ V }_{ 2 }-1.149{ V }_{ 2 } }{ 1.199{ V }_{ 2 } } \times 100$$
Loss %$$=\cfrac { 0.05 }{ 1.199 } \times 100$$
$$=5.995$$%
A body falling from a high Minaret travels $$40m$$ in the last $$2$$ seconds of its fall to ground. Height of Minaret in metres is
(take $$g=10m/{ s }^{ 2 }$$)
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0%
$$60$$
0%
$$45$$
0%
$$80$$
0%
$$50$$
Explanation
Taking the height of minaret is $$H$$ and time taken by body to fall from top to bottom be $$T$$.
$$\therefore H=(1/2)gT^2$$.....(1)
In last two second body travels a distance of 40m, henc ein $$(T-2)sec$$ body will travel $$(H-40)m$$.
$$ (H-40) = (1/2) g (T-2)^2$$.......(2)
$$\therefore $$ solving (1) and (2),
$$T=3sec , H=45m$$
Hence option $$B$$ is correct.
The speed of a car is reduced from 90 km/hr to 36 km/hr in 5 s. What is the distance travelled by the car during this time interval.
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0%
87.5 m
0%
90 m
0%
50 m
0%
100 m
Explanation
$$ velocity (u) = 90km/h$$.
$$= 25m/s$$
$$final velocity (v) = 36 km/h$$
$$= 10m/s$$
$$time(t) = 5 s$$.
acceleration (a)= $$\dfrac{v-u}{t}$$
= $$\dfrac{10-25}{5}$$
= $$\dfrac{-15}{5}$$
= $$-3m/s^2$$
let,
Using the third equation of motion
$$2as =v^2 - u^2$$
$$2 (-3)(s) = (10)^2 -(25)^2$$
$$-6s = 100 - 625$$
$$- 6s = - 525$$
$$6s = 525$$
$$s = \dfrac{525}{6}$$
$$= 87.5m$$
A particle is found to be at rest when seen from a frame $$S_{1}$$ and moving with a constant velocity when seen from another frame $$S_{2}$$. Mark out the possible options.
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Both the frames are inertal
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$$S_{1}$$ is inertial and $$S_{2}$$ is noninertial
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$$S_{1}$$ is noninertial and $$S_{2}$$ is inerital
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none of these
A particle is thrown upwards from ground. It experiences a constant resistance force which can produce retardation $$2{ m/s }^{ 2 }$$. The ratio of time of ascent to the time of descent is (g=$$10{ m/s }^{ 2 }$$)
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0%
1:1
0%
$$\sqrt { \dfrac { 2 }{ 3 } }$$
0%
$$\dfrac { 2 }{ 3 }$$
0%
$$\sqrt { \dfrac { 3 }{ 2 } }$$
Explanation
Let the maximum height attained by the particle be $$H$$.
Time taken by it to reach maximum height $$T = \sqrt{\dfrac{2H}{g_{eff}}}$$
During upward motion :
The particle experiences acceleration due to gravity $$10 \ m/s^2$$ in downward direction and also, resistive retardation $$2 \ m/s^2$$ in downward direction.
So, effective acceleration $$g_{eff} = 10+2 = 12 \ m/s^2$$
Thus, time of ascent $$t_a = \sqrt{\dfrac{2H}{12}}$$
During downward motion :
The particle experiences acceleration due to gravity $$10 \ m/s^2$$ in downward direction and also, resistive retardation $$2 \ m/s^2$$ in upward direction.
So, effective acceleration $$g_{eff} = 10-2 = 8 \ m/s^2$$
Thus, time of descent $$t_d = \sqrt{\dfrac{2H}{8}}$$
Thus $$\dfrac{t_a}{t_d} = \sqrt{\dfrac{2}{3}}$$
Correct answer is option B.
Rana moves with uniform velocity on a bike. He throws a stone in air, the stone falls:
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0%
Back in his hands
0%
In front of him
0%
At the back of him
0%
Cannot be predicted
Explanation
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A car and bike start racing in a straight line. The distance of finish line from starting line is 100m. The minimum acceleration of car to win, if it accelerates uniformly starting from rest and the bike moves with a constant velocity of 10 m/s, is
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0%
$$0.5 m/s^2$$
0%
$$2 m/s^2$$
0%
$$1m/s^2$$
0%
$$3 m/s^2$$
Explanation
For bike :
Speed $$v = 10 \ m/s$$
Distance to be covered $$d = 100 \ m$$
Time taken $$t = \dfrac{d}{v} = \dfrac{100}{10} = 10 \ s$$
For car to win the race :
Time $$t = 10 \ s$$
Initial velocity $$u = 0$$
Using $$S = ut+\dfrac{1}{2}at^2$$
Or $$100 = 0+\dfrac{1}{2}a(10)^2$$
$$\implies \ a = 2 \ m/s^2$$
A body is thrown horizontally from the top of a tower of height $$5\ m$$. It touches the ground at a distance of $$10\ m$$ from the foot of the tower. The initial velocity of the body is $$\left( g=10\ m{ s }^{ -2 } \right) $$
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0%
$$2.5\ { ms }^{ -1 }$$
0%
$$5\ { ms }^{ -1 }$$
0%
$$10\ { ms }^{ -1 }$$
0%
$$20\ { ms }^{ -1 }$$
Explanation
Height of the tower $$H= 5 \ m$$
Time of flight $$T = \sqrt{\dfrac{2H}{g}} = \sqrt{\dfrac{2\times 5}{10}} = 1 \ s$$
Range of projectile is $$R=10\ m$$
Initial velocity of projectile $$u = \dfrac{R}{T} = \dfrac{10}{1} = 10 \ m/s$$
Two bodies begin a free fall from the same height at a time interval of $$N$$ seconds. If vertical separation between the two bodies is $$1$$ m after n seconds from the start of the first body, then n is equal to
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0%
$$\sqrt{nN}$$
0%
$$\dfrac{1}{gN}$$
0%
$$\dfrac{1}{gN}+ \dfrac{N}{2}$$
0%
$$\dfrac{1}{gN}- \dfrac{N}{4}$$
Explanation
Let $$x_1$$ be the distance travelled by the first body.
Let $$x_2$$ be the distance travelled by the second body.
Using $$s=ut+\dfrac{1}{2}gt^2$$,
$$\Rightarrow -x_1= 0\times n -\dfrac{1}{2} gn^2$$ $$\Rightarrow x_1= \dfrac{1}{2}gn^2$$,
$$\Rightarrow -x_2= 0\times (n-N) -\dfrac{1}{2} g(n-N)^2$$ $$\Rightarrow x_2= \dfrac{1}{2}g(n-N)^2$$,
Given $$x_1-x_2=1=\dfrac{1}{2}gn^2-\dfrac{1}{2} g(n-N)^2$$ $$\Rightarrow n= \dfrac{1}{gN}+\dfrac{N}{2}$$
A car is moving with speed $$27km/h$$. The driver applied brakes as he approaches a circular turn on the road of radius $$80m$$ and his speed reduces at the constant rate of $$0.50m/s$$ every second. The magnitude of net acceleration is
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0%
$$20{ms}^{-2}$$
0%
$$0.86{ms}^{-2}$$
0%
$$100{ms}^{-2}$$
0%
None of these
Explanation
Speed $$v = 27 \ km/hr = 27\times \dfrac{5}{18} = 7.5\ m/s$$
Radius of circular turn $$r = 80 \ m/s$$
Radial acceleration $$a_r = \dfrac{v^2}{r} = \dfrac{7.5^2}{80} = 0.7 \ m/s^2$$
Tangential acceleration $$a_t = \dfrac{dV}{dt} = 0.50 \ m/s^2$$
Net acceleration $$a_{net} = \sqrt{a_r^2+a_t^2} = \sqrt{0.7^2+0.5^2} = 0.86 \ m/s^2$$
Correct answer is option B.
A juggler throws balls into air. He throws one when ever the previous one is at its highest point. If he throws n balls each second, the height to which each ball will rise is:
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0%
$$\dfrac{g}{2n^2}$$
0%
$$\dfrac{2g}{n^2}$$
0%
$$\dfrac{2g}{n}$$
0%
$$\dfrac{g}{4n^2}$$
Explanation
Juggler throws the ball into the air with ball having initial velocity $$= u$$
At the highest point of its path velocity will be zero as at maximum height velocity $$=0$$
So final velocity $$v=0$$
According to question juggler throws $$n$$ balls each second.
Time taken to reach highest position $$t= \dfrac 1n$$
Using equation of the motion we get:-
$$ v=u+at$$
$$ 0=u-\dfrac gn$$
$$u=\dfrac gn$$
Also,
Distance travelled in $$t=\dfrac 1n$$ secs:-
$$H= ut-\dfrac{gt^2}2$$
$$H=\dfrac g{n^2}- \dfrac g{2n^2}$$
Required height to which each ball rise, $$H= \dfrac g{2n^2}$$
A body $$A$$ starts from rest with an acceleration $$a_1$$. After $$2$$ seconds, another body $$B$$ starts from rest with an acceleration $$a_2$$. If they travel equal distances in the $$5^{th}$$ second, after the start of $$A$$, then the ratio $$a_1$$ : $$a_2$$ is equal to
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0%
$$5 : 9$$
0%
$$5 : 7$$
0%
$$9 : 5$$
0%
$$9 : 7$$
Explanation
Initial speed $$u = 0$$
Distance covered in nth second after starting from rest $$S = \dfrac{1}{2}(2n-1)$$
For A : $$n = 5$$
So, $$S_A = \dfrac{a_1}{2}(2\times 5-1) = \dfrac{9}{2}a_1$$
For B : $$n = 3$$
So, $$S_B = \dfrac{a_2}{2}(2\times 3-1) = \dfrac{5}{2}a_2$$
But $$S_A=S_B$$
Or $$\dfrac{9}{2}a_1 = \dfrac{5}{2}a_2$$
$$\implies \ a_1:a_2 = 5:9$$
A lift is coming from $$8^{th}$$ floor and is just about to reach $$4^{th}$$ floor. Taking ground floor as origin and positive direction upwards for all quantities, which one of the following is correct?
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x < 0, v < 0, a > 0
0%
x > 0, v < 0, a < 0
0%
x > 0, v < 0, a > 0
0%
x < 0, v < 0, a = 0
Explanation
$$v$$ is $$-ve$$ $$(v<0)$$
$$x(displacement)$$ is $$-ve$$ $$(x<0)$$
Lift is about to reach $$4^{th}$$ floor, here $$v=0\Longrightarrow v \quad is \quad \downarrow$$
$$\Longrightarrow a$$ is in opposite direction of $$v.$$
$$\Longrightarrow a=+ve.$$
A and B start walking towards each other from the opposite ends of a 15 km long straight road, at a speed of 5 km/hr and 7 km/hr respectively. How far apart will they be after one hour?
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0%
2 km
0%
3 km
0%
5 km
0%
7 km
Explanation
$$x_A=v_At=5\times 1=5km$$
$$x_B=v_Bt=7\times 1=7km$$
Total distance covered $$=12km$$
Distance between $$A$$ and $$B$$ at time $$t=1h$$
$$15-12=3km$$
On a long horizontally moving belt, a child runs to and fro with a speed 9 km $$h^{-1}$$ (with respect to the belt) between his father and mother located 50 m apart on the moving belt. The belt moves with a speed of 4 km $$h^{-1}$$. For an observer on a stationary platform, the speed of the child running in the direction of motion of the belt is
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0%
4 km $$h{-1}$$
0%
5 km $$h{-1}$$
0%
9 km $$h{-1}$$
0%
13 km $$h{-1}$$
Explanation
$${ V }_{ 1 }\rightarrow $$ Velocity of child wrt ground
$${ V }_{ 2 }\rightarrow $$ Velocity of belt wrt ground
$${ V }_{ child/belt }=9 \\{ V }_{ belt/g }=4$$
$${ V }_{ child/belt }={ V }_{ child/go }-{ V }_{ belt/g }$$
$$9={ V }_{ 1 }-4$$
$$\Rightarrow \boxed { { V }_{ 1 }=13km/hr } $$ [both are in same direction].
A tall cylinder is filled with viscous oil. A round pebble is dropped from the top with zero initial velocity. From the plots shown in figure, indicate the one that represents the velocity ($$v$$) of the pebble as a function of time ($$t$$).
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0%
0%
0%
0%
Explanation
When the pebble is dropped from the top of cylinder filled with viscous oil, pebble falls under gravity with constant acceleration, but as it is dropped it enters in oil and dragging force $$F = 6\eta \nu r v$$ due to viscosity of oil so acceleration decreases from g to zero i.e. velocity increases, but acceleration decreases, when acceleration decreased to zero, velocity becomes constant (terminal velocity). These conditions are verified in option ($$c$$).
A body covers a distance of 4 m in $$3^{rd}$$ second and 12 m in $$5^{th}$$ second. If the motion is uniformly accelerated, how far will it travel in the next 3 seconds?
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0%
10 m
0%
30 m
0%
40 m
0%
60 m
Explanation
We know that distance Travelled by uniformly accelerated body in $$n^{th}s$$ is given by
$$S_n=u+\dfrac{1}{2}a(2n-1),$$ where $$a$$ is the $$acceleration $$ of the body and $$u$$ is the initial velocity
it is given that $$S_n=4s$$ at $$n=3$$
$$\implies 4=u+\dfrac{5a}{2}$$ .............. $$eq(1)$$
$$S_n=12$$ at $$n=5s$$
$$\implies 12=u+\dfrac{9a}{2}$$ ............... $$eq(2)$$
solving $$eq(1)$$ and $$eq(2),$$ we get
$$a=4m/s^2$$ and $$u=-6m/s$$
Now we have to find velocity at $$t=5s$$
we know that $$v=u+at$$
$$\implies v=-6+(4\times 5)=14m/s$$
now distance travelled in next $$3s$$ can be calculated by
$$S=ut+\dfrac{1}{2}at^2$$ $$Note:$$ now $$u=14m/s$$
$$\implies S=14\times 3+\dfrac{1}{2}4(3^2)=60m $$
A body starts from rest and moves with constant acceleration for t s. It travels a distance $$x_1$$ in first half of time and $$x_2$$ in next half of time, then
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0%
$$x_2 = 3x_1$$
0%
$$x_2 = x_1$$
0%
$$x_2 = 4x_1$$
0%
$$x_2 = 2x_1$$
Explanation
$$x_1=0+\dfrac{1}{2}a{\left(\dfrac{t}{2}\right)}^2=\dfrac{at^2}{8}$$ $$[2^{nd}$$ equation of motion]
$$V_B=0+a\left (\dfrac{t}{2}\right)=\dfrac{at}{2}$$ $$[1^{st}$$ equation of motion]
$$x_2=\left (\dfrac{at}{2}\right)\dfrac{t}{2}+\dfrac{1}{2}(a) \left(\dfrac{t}{2}\right)^2$$
$$x_2=\dfrac{at^2}{4}+\dfrac{at^2}{8}=\dfrac{3at^2}{8}$$
$$\Longrightarrow x_2=3x_1$$
A rocket of initial mass $$6000\ kg$$ eject gases at a constant rate of $$16\ kg s^{-1}$$ with constant relative speed of $$11\ kms^{-1}$$. What is the acceleration of rocket one minute after the blast?
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0%
$$25 ms^{-2}$$
0%
$$50 ms^{-2}$$
0%
$$10 ms^{-2}$$
0%
$$35 ms^{-2}$$
Explanation
Acceleration of the rocket at any instant t is
$$a = \dfrac{v_r\dfrac{dm}{dt}}{M-t\dfrac{dm}{dt}}$$
Here, M=6000 kg, $$\dfrac{dm}{dt}=16 kg s^{-1}$$
$$v_r=11 km^{-1}= 11000 ms^{-1}$$, t=1 min = 60 s
$$\therefore a=[\dfrac{11000\times 16}{6000-60\times 16}]ms^{-2}=35 ms^{-2}$$
Which of the following statements is incorrect ?
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0%
In one dimension motion, the velocity and the acceleration of an object are always along the same line.
0%
In two or three dimensions, the angle between velocity and acceleration vectors may have any value between 0$$^{\circ}$$and 180$$^{\circ}$$
0%
The kinematic equations for uniform acceleration can be applied in case of uniform circular motion.
0%
The resultant acceleration of an object in circular motion is towards the centre only if the speed is constant.
Explanation
The kinematic equations for uniform acceleration do not apply in case of uniform circular motion because in this case the magnitude of acceleration is constant but its direction is changing.
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Practice Class 11 Engineering Physics Quiz Questions and Answers
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