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CBSE Questions for Class 11 Engineering Physics Work,Energy And Power Quiz 6 - MCQExams.com
CBSE
Class 11 Engineering Physics
Work,Energy And Power
Quiz 6
A force $$F_{x}$$ acts on a particle such that its position $$x$$ changes as shown in figure.
The work done by the particle as it moves from $$x = 0$$ to $$20\ m$$ is
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$$37.5\ J$$
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$$10\ J$$
0%
$$15\ J$$
0%
$$22.5\ J$$
0%
$$45\ J$$
Explanation
Key Concept As, we know, work done by a variable force, $$W_{n_{i}\rightarrow n_{t}}$$
$$=$$ area under the force $$-$$ displacement curve.
Work $$=$$ Area of $$\triangle ABF +$$ Area of $$\triangle BCEF +$$ Area of $$\triangle ECD$$
Work done by the particle as it moves from $$n = 0$$ to $$20\ m$$,
$$W = \dfrac {1}{2} \times 3\times 5 + 10\times 3 + \dfrac {1}{2} \times 5\times 3$$
$$\Rightarrow W = \dfrac {15}{2} + 30 + \dfrac {15}{2}$$
$$\Rightarrow W = 15 + 30 = 45\ J$$.
A particle is moving in a vertical circle. The tension in the string when passing through two platform at angles $$\displaystyle 30^o$$ and $$\displaystyle 60^o$$ vertical (lowest position) are $$T_1$$ and $$T_2$$ respectively
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$$\displaystyle T_1 = T_2$$
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$$\displaystyle T_2 > T_1$$
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$$\displaystyle T_1 > T_2$$
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Tension in the string always remain the same
Explanation
Tension, $$\displaystyle T= \frac{mv^2}{r} + mg cos \theta$$
For $$\displaystyle \theta =30^o$$
$$\displaystyle T_1= \frac{mv^2}{r} + mg cos 30^o$$
$$\displaystyle = \frac{mv^2}{r} + \frac{\sqrt{3}}{2} mg$$
For $$\displaystyle \theta =60^o$$
$$\displaystyle T_2= \frac{mv^2}{r} + mg cos 60^o$$
$$\therefore$$ $$\displaystyle \frac{mv^2}{r} + \frac{1}{2} mg$$
$$T_1> T_2$$
A stone of mass m is tied to a string and is moved in a vertical circle of radius r making n revolution per minute. The total tension in the string when the stone is at its lowest point is
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$$mg$$
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$$m(g+\pi n r^2)$$
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$$m(g+ n r)$$
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$$m(g+ \frac{\pi^2 n^2 r} {900})$$
Explanation
Tension in string when it reach lower-most point
$$\displaystyle T=mg+m \omega^2 r$$
$$\displaystyle =m(g+4 \pi^2 n^2 r)$$ [as $$\omega=2 \pi n$$]
$$\displaystyle = m \left( g + 4 \pi^2 \left( \frac{n}{60} \right)^2 r \right)$$
$$\displaystyle = m \left( g + \left( \frac{\pi^2 n^2 r}{900} \right) \right)$$
A particle moves along X-axis from $$x = 10$$ to $$x = 5\ cm$$ under the influence of force given by $$F = (7 - 2x + 3x^{2}) N$$. The work done in the process is
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$$70\ J$$
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$$270\ J$$
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$$35\ J$$
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$$135\ J$$
Explanation
The given force varies with distance.
Therefore, work done,
$$W = \displaystyle \int_{0}^{x = 5} F\cdot dx$$
$$W= \displaystyle \int_{0}^{5} (7 - 2x + 3x^{2})dx$$
$$W= \left [7x - \dfrac {2x^{2}}{2} + \dfrac {3x^{3}}{3} \right ]_{0}^{5}$$
$$W= [7 \times 5 - (5)^{2} + (5)^{3} - 0]$$
$$W= 135\ J$$.
A weightless thread can bear tension up to $$3.7$$kg wt. A stone of mass $$500$$g is tied to it and revolved in a circular path of radius $$4$$m in a vertical plane. If $$g=10m/s^2$$, then the maximum angular velocity of the stone will be :
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$$4rad/s$$
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$$2rad/s$$
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$$6rad/s$$
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none of these
Explanation
Maximum tension $$=\displaystyle\frac{mv^2}{r}+mg$$
$$3.7\times 10=\displaystyle\frac{0.5v^2}{4}+0.5\times 10$$
$$\Rightarrow v=16m/s$$
$$\therefore \omega =\displaystyle\frac{v}{r}=\frac{16}{4}=4rad/s$$.
A body of mass m is moving in a circle of radius r with a constant speed $$v$$. If a force $$\dfrac{mv^2}{r}$$ is acting on the body towards the centre, then what will be the work done by this force in moving the body over half the circumference of the circle?
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Zero
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$$\dfrac{mv^2}{r^2}$$
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$$\dfrac{mv^2}{r^2}\times \pi r$$
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$$\dfrac{\pi r^2}{mv^2}$$
Explanation
Work done by centripetal force is always zero. As the force is acting towards the centre so, the work done by this force in moving the body over half the circumference of the circle is zero.
A particle tied to a string describes a vertical circular motion of radius r continually. If it has a velocity $$\sqrt{3gr}$$ at the highest point, then the ratio of the respective tensions in the string holding it at the highest and lowest points is
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4 : 3
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5 : 4
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1 : 4
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3 : 2
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1 : 2
Explanation
Tension at highest point
$$T_H=\dfrac{mv^2}{r}-mg=3mg-mg=2mg$$
and tension at lowest point
$$T_L=\dfrac{mv^2}{r}+mg$$
Here, $$V^2_L=3gr+2g.2r=7gr$$
So, $$T_L=7mg+mg=8mg$$
Hence, $$\dfrac{T_H}{T_L}=\dfrac{2mg}{8mg}=\dfrac{1}{4}$$
A pendulum string of length $$l$$ is moved upto a horizontal position and released as shown in figure. If the mass of pendulum is $$m$$, then what is the minimum strength of the string that can withstand the tension as the pendulum passes through the position of equilibrium?
(neglect mass of the string and air resistance)
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$$mg$$
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$$2mg$$
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$$3mg$$
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$$5mg$$
Explanation
We have, $$\quad T-mg=\cfrac { m{ v }^{ 2 } }{ I } $$
$$\quad { v }^{ 2 }=2gI\Rightarrow T=mg+\cfrac { m2gI }{ I } =3mg$$
Two equal masses are attached to the two ends of a spring constant k. The masses are pulled out symmetrically to stretch the spring by a length x over its natural length. The work done by the spring on each mass during the above stretching is?
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$$\displaystyle\frac{1}{2}kx^2$$
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$$-\displaystyle\frac{1}{2}kx^2$$
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$$\displaystyle\frac{1}{4}kx^2$$
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$$-\displaystyle\frac{1}{4}kx^2$$
A simple pendulum with bob of mass $$m$$ and length $$x$$ is held in position at an angle $$1$$ and then angle $$2$$ with the vertical. When released from these positions, speeds with which it passes the lowest positions are $${v}_{1}$$ and $${v}_{2}$$ respectively. Then, $$\cfrac { { v }_{ 1 } }{ { v }_{ 2 } } $$ is
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$$\cfrac { 1-\cos { { \theta }_{ 1 } } }{ 1-\cos { { \theta }_{ 2 } } } $$
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$$\sqrt { \cfrac { 1-\cos { { \theta }_{ 1 } } }{ 1-\cos { { \theta }_{ 2 } } } } $$
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$$\sqrt { \cfrac { 2gx(1-\cos { { \theta }_{ 1 } } ) }{ 1-\cos { { \theta }_{ 2 } } } } $$
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$$\sqrt { \cfrac { 1-\cos { { \theta }_{ 1 } } }{ 2gx(1-\cos { { \theta }_{ 2 } } ) } } $$
Explanation
A simple pendulum with bob of mass m and length x is held in position at an angle 1 and then angle 2 with the vertical. When released from these positions, speeds with which it passes the lowest positions are $$v_1$$ and $$v_2$$ respectively.
Kinetic Energy = loss in PE
$$m v_1^2 /2 = m g L (1 - cos θ_1)$$
So,
Similarly $$m v_2^2 /2 = m g L (1 - cos θ_2)$$
Ratio $$\dfrac{v_1}{ v_2} =\sqrt{\dfrac{ (1 - cos θ_1)}{ (1 - cos θ_1)}}$$ .
When a rubber -band is stretched by a distance x,it exerts a restoring force of magnitude $$F= ax + bx^2$$ where a and b are constants. the work done is stretching the unstretched rubber band by L is :
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$$\dfrac {aL^{2}}{2}+\dfrac{bL^{3}}{3}$$
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$$\dfrac{1}{2} \begin {pmatrix} \dfrac{aL^2}{2} + \dfrac{bL^3}{3}\end {pmatrix} $$
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$$aL^2 + bL^3$$
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$$\dfrac {1}{2} ( aL^2 + bL^3)$$
A block of mass $$M$$ at the end of the string is whirled round a vertical circle of radius $$R$$. The critical speed of the block at the top of the swing is
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$${ \left( R/g \right) }^{ 1/2 }$$
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$$g/R$$
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$$M/Rg$$
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$${ \left( Rg \right) }^{ 1/2 }$$
Explanation
Using conservation of energy
Total mechanical energy at lowest point = Total mechanical energy at top
$$\dfrac{1}{2}m{v_{lowest}^2} = \dfrac{1}{2}m{v_{High}^2} + 2mgR$$
we know that $$ {v_{lowest}} = \sqrt{5Rg}$$
$$\therefore \dfrac{1}{2}(5Rg) = \dfrac{1}{2}{v_{high}^2} + 2gR$$
$${v_{high}} = \sqrt{Rg}$$
The mass of bucket full of water is 15 kg . It is being pulled up from a 15 m deep well. Due to a hole in the bucket 6 kg water flows out of the bucket at a uniform rate. The work done in drawing the bucket out of the well will be $$(g=10 m/s^2)-$$
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900 J
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1500 J
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1800 J
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2100 J
Explanation
$$\textbf{Step 1: Water in bucket at height x} $$
Mass of bucket full of water $$ = 15\ kg$$
Water goes out of bucket $$ = 6\ kg$$
Out flow of water per meter of pull $$ = \dfrac{6kg}{15m} $$
$$ = 0.4\ kg/m$$
Water in the bucket at height x
from bottom $$(m)$$
$$ = (15 - 0.4 x) kg$$
$$m = (15 - 0.4x)$$
$$\textbf{Step 2: Work done} $$
Work done in pulling the bucket by small distance $$(dx)$$
$$dW = mg\ dx $$
$$dW = (15 - 0.4 x)g\ dx $$
$$\textbf{Step 3: Integration}$$
For total work done integrate from $$x = 0$$ to $$x = 15\ m$$
$$W = \int^{15}_{0} (15 - 0.4 x) g\ x $$
$$\Rightarrow W = g \left [15 x - 0.4 \dfrac{x^2}{2} \right ]^{15}_{0} $$
$$\Rightarrow W = 10 (225 - 45) \quad$$
$$\Rightarrow W = 1800\ J$$
Option C is correct.
$$\textbf{Alternate Solution:} $$
Here work done by gravity is linear function of height $$(x)$$
So we can find average water leak in this case-
Average weight leak $$ = 3$$ g Kg - wt
So remaining $$ = (15 - 3)\ g = 12$$ g Kg - wt
Work done $$ = F \times d $$ $$ = 12\ g \times 15$$
$$ = 120 \times 15$$
$$W = 1800\ J$$
$$\textbf{Note:} $$
This method is only applicable for the linear function of height.
If work done by gravity is a non-linear function of $$x$$ then, don't use this approach
e.g. If $$W = mgx^3$$
Here we can not apply this method
If the force constant of a wire is $$K$$, the work done in increasing the length of the wire by/ is
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$$Kl/ 2$$
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$$Kl$$
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$$KI^{2}/2$$
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$$Kl^{2}$$
Explanation
Let initial length of the wire be L
Then final length of the wire will be L+l
Initial distance of COG of wire from the bottom $$=\dfrac{L}{2} $$
Final distance of COG of wire from the bottom $$=\dfrac{L+l}{2} $$
Displacement of the COG of the wire
$$s=\dfrac{L+l}{2} $$
$$-\dfrac{L}{2} $$
$$=\dfrac{l}{2} $$
Work done $$=Ks=K\dfrac{l} {2}$$
Hence correct answer is option $$A $$
The work done by external agent in stretching a spring of force constant $$k = 100\ N/cm$$ from deformation $$x_{1} = 10$$ to deformation $$x_{2} = 20\ cm$$.
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$$-150\ J$$
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$$50\ J$$
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$$150\ J$$
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None of these
Explanation
Work done by external agent =change in potential energy of spring
$$=\dfrac{1}{2}kx_2^2-\dfrac{1}{2}kx_1^2$$
$$=\dfrac{1}{2} (10000 N/m)(0.2)^2-\dfrac{1}{2}(10000)(0.1)^2$$
$$=5000(0.04-0.01)=150 J$$
A particle is projected so as to just move along a vertical circle of radius $$r$$ with the help of the massless string. The ratio of the tension in the string when the particle is at the lowest and highest point on the circle is?
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$$1$$
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Finite but large
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Zero
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Infinite
Explanation
$$T-mg\cos\theta=\cfrac{mV^2}{r}$$
At lowest point $$A$$:
$$T_A=\cfrac{mV_A^2}{r}+mg\cos (0^o)$$
$$\implies T_A=\cfrac{mV_A^2}{r}+mg$$
At highest point $$B$$:
$$T_B=0$$
$$\therefore \cfrac{T_A}{T_B}=\cfrac{({mV_A^2/r})+mg}{0}$$
$$\implies \cfrac{T_A}{T_B}=\infty$$
The diagram below shows the path taken by a ball when Sundram kicks it. The potential energy of the ball is highest at ______________
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P
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Q
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R
0%
S
When the speed of a body is doubled, its kinetic energy becomes
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Double
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Half
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Quadruple
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One-fourth
Explanation
The kinetic energy of a body is given by:
$$KE=\dfrac{1}{2}mv^2$$
Hence, Kinetic energy depends on the velocity as:
$$KE\propto v^2$$
So, if we double the velocity, then $$KE$$ becomes four times.
Hence, option C is correct.
A force acts on a $$3 g$$ particle in such a way that position of the particle as a function of time is given by $$x=3t-4t^2+t^3$$, where x is in metre and t is in sec. The work done during the first $$4s$$ is
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570 mJ
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450 mJ
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490 mJ
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528 mJ
Explanation
$$\dfrac{dx}{dt}=3-8t+3t^2$$
$$dx=(3-8t+3t^2)dt$$
$$\dfrac{d^2}{dx^2}=a=-8+6t$$
Force $$F=ma=m(-8+6t)$$
Work done $$F.dx$$
$$F.(3-8t+3t^2)dt$$
$$\int _{ o }^{ w }{ dw\quad =\int _{ 0 }^{ 4 }{ m\left( -8+6t \right) \left( 3-8t+3{ t }^{ 2 } \right) } } dt\\ w=m\int _{ 0 }^{ 4 }{ \left( -24+82t-72{ t }^{ 2 }+18{ t }^{ 3 }\quad \right) dt } \\ =m\overset { 4 }{ \underset { 0 }{ \left[ -24t+41{ t }^{ 2 }-24{ t }^{ 3 }+\dfrac { 3 }{ 2 } { t }^{ 4 } \right] } } \\ =m(176)\\ =3\times { 10 }^{ -3 }\times 176\\ =528\quad mJ$$
A particle is tied to one end of a light inextensible string and is moved in a vertical circle, the other end of the string is fixed at the centre. Then for a complete motion in a circle, which is correct.
(air resistance is negligible).
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Acceleration of the particle is directed towards the centre
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Total mechanical energy of the particle and earth remains constant
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Tension in the string remains constant
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Acceleration of the particle remains constant
Explanation
At any time force acting on particle vary and hence acceleration (net) will have different direction at different times. Tension also changes and its minimum at top point. Magnitude of acceleration also varies.
Considering earth and particle as a system and no external force on system is acting, total mechanical energy will be conserved.
A triangular block ABC of mass m and sides 2a lies on a smooth horizontal plane as shown in the figure. Three point masses of mass m each strike the block at A, B and C with speeds v as shown. After the collision, the particles come to rest. Then the angular velocity acquired by the triangular block is (I is the moment of inertia of the triangular block about G, perpendicular to the plane of the block)
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$$\displaystyle 2mva \frac{(1 + \sqrt{3})}{\sqrt{3} l}$$ clockwise
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$$\displaystyle \frac{2mva}{ l}$$ clockwise
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$$\displaystyle \frac{2 \sqrt{3} mva}{ l}$$ clockwise
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None of these.
Explanation
$$h=\dfrac { 2 }{ 3 } \sqrt { 3 } a\dfrac { 2 }{ \sqrt { 3 } } a\\ I\omega =3mvh\\ =2\sqrt { 3 } mva\\ \omega =\dfrac { 2\sqrt { 3 } mva }{ l } $$
A simple pendulum of length $$L$$ carries a bob of mass $$m$$. When the bob is at its lowest position, it is given the minimum horizontal speed necessary for it to move in a vertical circle about the point of suspension. When the stri
ng is horizontal the net force on the bob is:
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$$\sqrt { 10 } mg$$
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$$\sqrt { 5 } mg$$
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$$4mg$$
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$$1mg$$
Explanation
V=minimum horizontal speed= $$\sqrt { 5Lg } $$
using energy conservation,
$$\frac { 1 }{ 2 } m{ v }^{ 2 }-\frac { 1 }{ 2 } m{ u }^{ 2 }=mgL\\ \therefore \quad \frac { 1 }{ 2 } m{ u }^{ 2 }=\frac { 1 }{ 2 } m{ v }^{ 2 }-mgL\\ =\frac { 1 }{ 2 } m{ 5gL }-mgL\\ \therefore \quad { v }^{ 2 }=\frac { 3gL }{ 2 } \times 2=3gL\\ T=\frac { m{ v }^{ 2 } }{ L } =3mg\\ \therefore \quad force=\quad \sqrt { { \left( 3mg \right) }^{ 2 }+{ \left( mg \right) }^{ 2 } } \quad \\ =\sqrt { 10 } mg$$
Which of the following statements is incorrect?
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Kinetic energy may be zero, positive or negative
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Power, energy and work are all scalars
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Potential energy may be zero, positive or negative
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Ballistic pendulum is a device for measuring the speed of bullets
Explanation
The kinetic energy of a body of mass
m
m
which is moving with velocity
v
v
is
$$ K = \dfrac{1}{2}mv^2$$
Mass is always positive and if the velocity is either positive or negative, the kinetic energy is always positive due to the square of velocity.
When a rubber-band is stretched by a distance $$x$$, it exerts a restoring force of magnitude $$F=ax+b{x}^{2}$$ where $$a$$ and $$b$$ are constants.The work done in stretching the unstretched rubber band by $$L$$ is:
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$$\cfrac { a{ L }^{ 2 } }{ 2 } +\cfrac { { bL }^{ 3 } }{ 3} $$
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$$\cfrac { 1 }{ 2 } \left( \cfrac { a{ L }^{ 2 } }{ 2 } +\cfrac { { bL }^{ 3 } }{ 2 } \right) $$
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$$a{ L }^{ 2 }+{ bL }^{ 3 }$$
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$$\cfrac { 1 }{ 2 } \left( a{ L }^{ 2 }+{ bL }^{ 3 } \right) $$
Explanation
Work done by streching dx length is,
$$dw=Fdx$$
$$=(ax+bx^2)dx$$
$$w=\int _{ 0 }^{ L }{ { \left( ax+b{ x }^{ 2 } \right) }dx } \\ =\overset { L }{ \underset { 0 }{ \left[ \dfrac { { ax }^{ 2 } }{ 2 } +\dfrac { { bx }^{ 3 } }{ 3 } \right] } } \\ =\dfrac { { aL }^{ 2 } }{ 2 } +\dfrac { { bL }^{ 3 } }{ 3 } $$
Which of the following is not an example of potential energy?
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A vibrating pendulum at its maximum displacement from the mean position
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A body at rest at some height from the ground
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A wound clock-spring
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A vibrating pendulum when it is just passing through the mean position
Explanation
Whenever a body is at some height from the mean (base) position or at a state of extension (or contraction), the body will have some potential energy. But when the pendulum is passing the mean position, the height from the base is 0. Hence a vibrating pendulum at the mean position has no potential energy.
A body of mass $$0.5\ kg$$ travels in a straight line with velocity $$v = ax^{1/2}$$ where $$a = 4 m^{}s^{-2}$$. The work done by the net force during its displacement from $$x = 0$$ to $$x = 2\ m$$ is
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0%
$$1.5\ J$$
0%
$$50\ J$$
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$$10\ J$$
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None of these
A force $$F$$ acting on an object varies with distance $$x$$ as shown in the figure. The work done by the force in moving the object from $$x = 0$$ and $$x = 20\ m$$ is
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0%
$$500\ J$$
0%
$$1000\ J$$
0%
$$1500\ J$$
0%
$$2000\ J$$
Explanation
Area under force displacement curve is the work done in that interval
Area under the given figure$$=$$ Area of surface$$+$$ Area of triangle.
Work done$$=10\times 100+\cfrac { 1 }{ 2 } \times 10\times 100$$
$$=1000+500\\ =1500J$$
When spring is cut into parts its effective spring constant of each part decreases.
State whether True/False?
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True
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False
Explanation
Spring constant of spring is inversely proportional to the length of the spring.
So when spring cut into parts its effective spring constant of each part would increase.
A body of mass starts moving from rest along x-axis so that its velocity varies as $$v = a\sqrt {s}$$ where $$a$$ is a constant and $$s$$ is the distance covered by the body. The total work done by all the forces acting on the body in the first seconds after the start of the motion is:
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$$\dfrac {1}{8} ma^{4}t^{2}$$
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$$4\ ma^{4}t^{2}$$
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$$8 ma^{4}t^{2}$$
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$$\dfrac {1}{4}ma^{4}t^{2}$$
Explanation
Velocity of body $$v = a \sqrt s$$
acceleration $$a' = \dfrac{dv}{dt} = \dfrac{a}{2 \sqrt s} \dfrac{ds}{dt} = \dfrac{a^2}{2}$$
displacement $$s' = \dfrac{1}{2} a' t^2 = \dfrac{1}{2} \dfrac{a^2}{2} t^2$$
Work done $$W = force \times displacement$$
$$W = ma' \times s' = \dfrac{1}{8} ma^4 t^2$$
A body of mass $$0.5\ kg$$ travels in a straight line with velocity $$v = ax^{3/2}$$ where $$a = 5\ m^{-1/2}s^{-1}$$. The work done by the net force during its displacement from $$x = 0$$ to $$x = 2m$$ is
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$$1.5\ J$$
0%
$$50\ J$$
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$$10\ J$$
0%
$$100\ J$$
Explanation
Given: $$m = 0.5\ kg, v = kx^{3/2}$$ where, $$k = 5\ m^{-1/2}s^{-1}$$
Acceleration, $$a = \dfrac {dv}{dt} = \dfrac {dv}{dx}\dfrac {dx}{dt} = v\dfrac {dv}{dx} \left (\because v = \dfrac {dx}{dt}\right )$$
As $$v^{2} = k^{2}x^{3}$$
Diferentiating both sides with respect to $$x$$, we get
$$2v \dfrac {dv}{dx} = 3k^{2}x^{2}$$
$$\therefore Acceleration, a = \dfrac {3}{2} k^{2}x^{2}$$
Force, $$F = Mass\times Acceleration = \dfrac {3}{2} mk^{2}x^{2}$$
Work done, $$W = \int Fdx = \int_{0}^{2}\dfrac {3}{2} mk^{2}x^{2}dx$$
$$W = \dfrac {3}{2} mk^{2}\left [\dfrac {x^{3}}{3}\right ]_{0}^{2} = \dfrac {3}{6} \times 0.5\times 5^{2} \times [2^{3} - 0] = 50\ J$$.
A man revolves a stone of mass m tied to the end of a string in a vertical circle of radius R, The net force at the lowest and height points of the circle directed vertical downwards are
Here $$T_1, T_2$$ and $$v_1, v_2$$ denote the tension in the string and the speed of the stone at the lowest and highest points, respectively.
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Lowest point: $$mg - T_1$$ Highest point : $$mg + T_2$$
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Lowest point: $$mg + T_1$$ Highest point : $$mg - T_2$$
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Lowest point: $$mg + T_1 -\frac{mv^2}{R}$$ Highest point: $$mg - T_2 +\frac{mv^2}{r}$$
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Lowest point: $$mg - T_1 -\frac{mv^2}{R}$$ Highest point: $$mg + T +\frac{mv^2}{r}$$
A force F is related to the position of a particle by the relation $$ F =(10x^2)N$$. The work done by the force when the particle moves from x=2m to x=4 m is
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$$\dfrac{56}{3}J$$
0%
$$560 J$$
0%
$$\dfrac {560}{3}J$$
0%
$$\dfrac{3}{560} J$$
Explanation
$$F=10 x^{2}N$$
We know that work done $$=\displaystyle\int_{x_{1}}^{x_{2}}F.dx$$
$$\omega=\displaystyle\int_{2}^{4}10 x^{2}dx$$
$$\Rightarrow \omega=10\displaystyle\int_{2}^{4}x^{2}dx$$
$$=10\left[\dfrac{x^{3}}{3}\right]_{2}^{4}$$
$$=10\left[\dfrac{64}{3}-\dfrac{8}{3}\right]=\dfrac{560}{3}\ J$$
Two wires $$AC$$ and $$BC$$ are tied at $$C$$ of a small sphere of mass $$5\ kg$$, which revolves at a constant speed $$v$$ in the horizontal plane with the speed $$v$$ of radius $$1.6\ m$$. Find the minimum value of $$v$$.
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$$4\ m{s}^{-1}$$
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$$2\ m{s}^{-1}$$
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$$2.5\ m{s}^{-1}$$
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None of these
Explanation
From $$FBD$$
$$T_1\cos 30^0+T_2\cos 45^0=mg$$
$$T_1\sin 30^0+T_2\sin 45^0=\dfrac{mv^2}{r}$$
$$T_1\sin 30^0-T_1\cos 30^0=\dfrac{mv^2}{R}-mg$$ $$\because\{\sin45^0=\cos 45^0\}$$
$$T_1=\dfrac{\dfrac{mv^2}{R}-mg}{\left[\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}\right]}$$
as it a string tension, So it can't be negative
$$T_1 > 0$$ i.e $$\dfrac{mv^2}{R}-mg \ge 0$$
$$\dfrac{v_2}{r}-g \ge 0$$
$$v\ge \sqrt{gR}$$
$$V_{min}\sqrt{gr}=\sqrt{10\times 1.6}=\sqrt{16}=4ms^{-1}$$
Hence (A) is the correct options.
The velocity of a body moving in a vertical circle of radius r is $$\sqrt{7gr}$$ at the lowest point of the circle. What is the ratio of maximum and minimum tension?
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$$4: 1$$
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$$\sqrt 7:1$$
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$$3: 1$$
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$$2: 1$$
Explanation
Since $$V_{b} >\sqrt{5 gr}\rightarrow$$ It will complete vertical circular
$$\downarrow$$
velocity at bottom
Thus $$T_{bottom}-T_{top}=6\ mg$$
$$T_{max}-T_{min}=6\ mg ...... (1)$$
Now,
$$\Rightarrow T_{max}-mg=m(a_{c})$$
$$T_{max}-mg =\dfrac{mv^{2}}{r}$$
$$T_{max}=\dfrac{m}{r}(7 gr)+mg$$
$$T_{max}=8\ mg$$
From $$(1)$$
$$T_{min}=2\ mg$$
Thus $$\dfrac{T_{max}}{T_{min}}=4:1$$
two blocks of masses $$m_1 = 2 kg$$ and $$m_2 = 4 kg$$ are moving in the same direction with speeds $$\nu_1 = 6 m/s$$ and $$\nu_2 = 3 m/s$$, respectively on a frictioneless surface as shown in the figure. An ideal spring with spring constant $$k = 30000 N/m$$ is attached to the back side of $$m_2$$. Then the maximum compression of the spring after collision will be:
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$$0.06 m$$
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$$0.04 m$$
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$$0.02 m$$
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none of these
Explanation
$$m_{1}=2\ kg ,m_{2}=4\ kg$$
$$v_{1}=6\ m/s, v_{2}=3\ m/s$$
$$k=30000\ N/m$$
From momentum conesrvation
$$2\times 6+5\times 3=2v_{1}+5v_{2}$$
$$2v_{1}+5v_{2}=27$$
$$v_{1}=v_{2}=v=\dfrac{27}{7}m/s$$ for max compression
From Energy conservation
$$\dfrac{1}{2}mv^{2}$$
$$\dfrac{1}{2}m_{1}v_{1}^{2}+\dfrac{1}{2}m_{2}v_{2}^{2}=\dfrac{1}{2}m_{2}v^{2}+\dfrac{1}{2}m_{2}v^{2}+\dfrac{1}{2}kx^{2}$$
$$\dfrac{1}{2}\times 2\times 6\times 6+\dfrac{1}{2}\times 4\times 3\times 3=\dfrac{1}{2}\times 2\times \dfrac{27\times 27}{7\times 7}+\dfrac{1}{2}\times \dfrac{4\times 27\times 27}{7\times 7}+\dfrac{1}{2}\times 30,000\times x^{2}$$
$$36+18=14.88+29.75+15000 x^{2}$$
$$54-44.63=15000 x^{2}\Rightarrow 937=15000 x^{2}$$
$$x=0.02\ m$$
A particle of mass 1 kg is suspended by means of a string of length L=2 m. The string makes $$6/\pi$$ rps around a vertical axis through the fixed end. The tension in the string is
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72 N
0%
36 N
0%
288 N
0%
10 N
Explanation
The tension is the centripetal force = $$mw^2R$$
The angular velocity is w=$$(6/\pi)(2 \pi) =12$$ rads/s
Substituting we get, Tension = $$1 \times 144 \times 2 = 288 N$$
Thus the correct options is (c)
A 0.25kg ball attached to a 1.5 m rope moves with a constant speed of 15 m/s around a vertical circle. Calculate the tension force on the rope at the middle of the circle:
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37.5 N
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137.5 N
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2.5 N
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25 N
Explanation
The centripetal force acts along the direction of the tension and hence $$T=mv^2/R = 0.25 \times 15^2/1.5=37.5 N$$
A mass attached to a string that is itself attached to the ceiling swings back and forth. If the bob is observed to be moving upward at a given instance, as shown to the right, which arrow best depicts the direction of the net force acting on the bob at that instant
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A
0%
B
0%
C
0%
D
Explanation
Two forces act on the bob. (i) is the tension in the string and the (ii) mg sin $$\theta$$, which will be tangential to the path. The resultant of both the forces will be along vector C
Thus the correct option is (c)
A time-varying force $$F=6t-2{ t }^{ 2 } N$$, at $$t=0$$ starts acting on a body of mass $$2 kg$$ initially at rest, where t is in second. The force is withdrawn just at the instant when the body comes to rest again. We can see that at $$t=0$$,the force $$F=0$$. Now answer the following:
Mark the correct statement:
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Velocity of the body is maximum when force acting on the body is maximum for the first time.
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The velocity of the body becomes maximum when force acting on the body becomes zero again
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When force becomes zero again, velocity of the body also becomes zero at that instant.
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All of the avove
A force $$F = -K(x\hat {i} + y\hat {j})$$ (where $$K$$ is a positive constant) acts on a particle moving in the $$x-y$$ plane. Starting from the origin, the particle is taken along the positive $$x-$$ axis to the point $$(a, 0)$$ and then to the point $$(a, a)$$. The total work done by the force $$\vec {F}$$ on the particle is
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$$-2Ka^{2}$$
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$$2Ka^{2}$$
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$$-Ka^{2}$$
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$$Ka^{2}$$
Explanation
For motion of the particle from $$(0,0)$$ to $$(a,0)$$
$$\vec{F}=-K(0\hat{i}+a\hat{j})\Rightarrow \vec{F}=-Ka\hat{j}$$
Displacement $$\vec{r}=(a\hat{i}+0\hat{j})-(0\hat{i}+0\hat{j})=a\hat{i}$$
So work done from $$(0,0)$$ to $$(a,0)$$ is given by
$$W=\vec{f}.\vec{r}=-Ka\hat{j}.a\hat{i}=0$$
For motion $$(a,0)$$ to $$(a,a)$$
$$\vec{F}=-K(a\hat{i}+a\hat{j})$$ and displacement
$$\vec{r}=(a\hat{i}+a\hat{j})-(a\hat{i}+0\hat{j})=a\hat{j}$$
So work done from $$(a,0)$$ to $$(a,a)$$ $$W=\vec{F}.\vec{r}$$
$$=-K(a\hat{i}+a\hat{j}).a\hat{j}=-Ka^{2}$$
So total work done $$=-Ka^{2}$$
A spring of spring constant $$5\times {10}^{3}N/m$$ is stretched initially by $$5cm$$ from unstretched position. Then the work required to stretch is further by another $$5cm$$ is then
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$$12.50N-m$$
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$$18.75N-m$$
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$$25.00N-m$$
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$$6.25N-m$$
Explanation
$$\Delta W=\cfrac { 1 }{ 2 } \left( 5\times { 10 }^{ 3 } \right) \left[ { (0.1) }^{ 2 }-{ (0.5) }^{ 2 } \right] =18.75$$
A mass of $$M\ kg$$ is suspended by a weightless string. The horizontal force that is required to displace it unitl the string makes an angle of $${45}^{o}$$ with the initial vertical direction is
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$$Mg(\sqrt{2}-1)$$
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$$Mg(\sqrt{2}+1)$$
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$$Mg\sqrt {2}$$
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$$\cfrac { Mg }{ \sqrt { 2 } } $$
Explanation
Work done by $$F$$
$$F=\cfrac { l }{ \sqrt { 2 } } =mg\left( l-\cfrac { l }{ \sqrt { 2 } } \right) $$
$$F=Mg\left( \sqrt { 2 } -1 \right) $$
A thin uniform rod of mass $$m$$ and length $$l$$ is hinged at the lower end of a level floor and stands vertically. It is now allowed to fall, then its upper and will strike the floor with a velocity given by
(A)$$\sqrt { mgl }$$(B) $$\sqrt { 3gl }$$
(c)$$\sqrt { 5gl }$$ (D) $$\sqrt { 2gl }$$
Sol.
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A
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B
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C
0%
D
Explanation
Initially rod stand vertically
Its potential energy $$=mg=\dfrac{l}{2}$$
When it striker the floor, its potential energy will convert into rotational kinetic energy,
$$mg\left(\dfrac{1}{2}\right)=\dfrac{1}{2}I\omega^2$$
Where, $$I=\dfrac{ml^2}{3}=M.I$$ of rod about point A
$$\therefore mg\left(\dfrac{l}{2}\right)=\dfrac{l}{2}\left(\dfrac{ml^2}{3}\right)\left(\dfrac{V_B}{l}\right)^2$$
$$V_V=\sqrt{3gl}$$
In an elastic collision between two particles
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net kinetic force is zero
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the kinetic energy of the system before collision is equal to the kinetic of the system after collision
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linear momentum of system before collision = linear momentum after collision
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the total energy of the system is never conserved
Explanation
We know,
Net Kinetic force involved in the collision is zero in case of elastic collision.
And as there is no net Kinetic force, net work done during collision=0,
And hence $$\Delta KE=0$$ during collision.
Hence there is no change in KE before and after collision.
And as there is no external force involved in collision,
Momentum of the colliding system is conserved.
And hence total momentum before collision=total momentum after collision.
Note; momentum is conserved irrespective of elastic/inelastic collision.
Option $$\textbf{A,B,C}$$ is correct answer.
The work done by a force is equal to :
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the area under $$f\left( x \right) $$ vs $$x$$ curve and $$x$$ axis
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half the area under $$f\left( x \right) $$ vs $$x$$ curve and $$x$$ axis
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the area under $$f\left( x \right) $$ vs $$x$$ curve and $$F$$ axis
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half the area under $$f\left( x \right) $$ vs $$x$$ curve and $$F$$ axis
Explanation
he correct option is A.
Therefore,
$$W=\int Fdx$$
When a rubber-bank is stretched by a distance $$x$$, it exerts a restoring force of magnitude $$F=ax+b{x}^{2}$$ where $$a$$ and $$b$$ are constants. The work done in stretching the unstretched rubber band by $$L$$ is:
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$$\cfrac { a{ L }^{ 2 } }{ 2 } +\cfrac { b{ L }^{ 3 } }{ 3 } $$
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$$\cfrac { 1 }{ 2 } \left( \cfrac { a{ L }^{ 2 } }{ 2 } +\cfrac { b{ L }^{ 3 } }{ 3 } \right) $$
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$$a{ L }^{ 2 }+b{ L }^{ 3 }$$
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$$\cfrac { 1 }{ 2 } \left( a{ L }^{ 2 }+b{ L }^{ 3 } \right) $$
Explanation
$$F=ax+b{ x }^{ 2 }$$
$$dw=Fdx$$
$$W=\int _{ 0 }^{ L }{ \left( ax+b{ x }^{ 2 } \right) } dx\quad $$
$$W=\cfrac { a{ L }^{ 2 } }{ 2 } +\cfrac { b{ L }^{ 3 } }{ 3 } $$
A particle moves along $$X-$$axis from $$x=0$$ to $$x=1\ m$$ under the influence of a force given by $$F=3x^{2}+2x-10$$. Work done in the process is:
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$$+4\ J$$
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$$-4\ J$$
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$$+8\ J$$
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$$-8\ J$$
Explanation
Given,
$$F=3x^2+2x-10$$
Work done $$=\int { F\cdot ds }$$
$$ =\int _{ 0 }^{ 1 }{ \left( { 3x }^{ 2 }+2x-10 \right) dx } $$
$$=\left[{ \frac { { 3x }^{ 3 } }{ 3 } +\frac { { 2x }^{ 2 } }{ 2 } -10x }\right]_{ 0 }^{ 1 }$$
$$=(1)^3+(1)^2-10-0$$
$$=2-10$$
$$=-8J$$
A body of mass $$1kg$$ thrown upwards with a velocity of $$10m/s$$ comes to rest (momentarily) after moving up by $$4m$$. The work done by air drag in this process is (Take $$g=10m/{ s }^{ 2 }$$)
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$$-20J$$
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$$-10J$$
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$$-30J$$
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$$0J$$
$$4\ J$$ of work is required to stretch a spring through $$10\ cm$$ beyond its unstreched length. The extra work required to stretch it through additional $$10\ cm$$ shall be
Report Question
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$$4\ J$$
0%
$$8\ J$$
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$$12\ J$$
0%
$$16\ J$$
Explanation
W = 4J
X = 10cm = 0.1 m = $$10^{-1}$$
By Hook's law,
W = $$\frac{1}{2}$$k$$x^2$$_________eq1
4 =
$$\frac{1}{2}$$ x
k x 0.1 x 0.1
8 = k x $$10^{-2}$$
k = 800 N$$m^{-1}$$
Additional 10cm means 0.1 + 0.1 i.e 0.2
Therefore again by same law,
$$w_1$$
=
$$\frac{1}{2}$$ x
800 x 4 x $$10^{-2}$$
$$w_1$$ = $$16 J$$
AS we know,
w =
$$w_1$$ - W
w = $$12 J$$
A particle moves along $$y=\sqrt { 1-{ x }^{ 2 } } $$ betweem the points $$(0,-1)m$$ and $$(0,1)m$$ under the influence of a force $$\overrightarrow { F } =\left( { y }^{ 2 }\hat { i } +{ x }^{ 2 }\hat { j } \right) N$$. Then
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the particle is moving along a semi-ellipse
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the particle is moving along a semicircle
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work done on the particle by $$\overrightarrow { F } $$ is $$(3/4)J$$
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work done on the particle by $$\overrightarrow { F } $$ is $$(4/3)J$$
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Practice Class 11 Engineering Physics Quiz Questions and Answers
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