MCQ Questions for Class 11 Engineering Physics Motion In A Straight Line Quiz 15

A ball is dropped from a bridge of

$122.5$ metre above a river. After the ball has been falling for two seconds, a second ball is thrown straight down after it. Initial velocity of the second ball so that both hit the water at the same time is

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$49\, m/s$
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$55.5\, m/s$
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$26.1\, m/s$
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$9.8\, m/s$

A car is moving with a speed of

$10\,m/s$ on a circular path of radius

$25m$ . Driver of car applies the brakes producing a uniform deceleration of

$3\,m/s^2$ . Then,

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the centripetal acceleration of car just after applying the brake is $4\,m/s^2$
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the acceleration just after applying the brake is $5\,m/s^2$
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the acceleration is directed towards the center just after applying the brake
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the angle between acceleration and velocity just after applying the brake is $127^o$

A particle is moving along the x-axis whose acceleration is given a=3x-4, where x is the location of the particle. At t=0, the particle is at rest at x=4/3 m. The distance travelled by the particle is 5 s is

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zero
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42 m
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Infinite
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None of these

A stone falls freely under gravity. It covers distances

$h _ { 1 } , h _ { 2 }$ and

$h _ { 3 }$ in the first

$5$ seconds, the next

$5$ seconds and the next

$5$ seconds respectively. The relation between

$h _ { 1 } , h _ { 2 }$ and

$h _ { 3 }$ is

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$h _ { 1 } = \dfrac { h _ { 2 } } { 3 } = \dfrac { h _ { 3 } } { 5 }$
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$h _ { 2 } = 3 h _ { 1 } \text { and } h _ { 3 } = 3 h _ { 2 }$
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$h _ { 1 } = h _ { 2 } = h _ { 3 }$
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$h _ { 1 } = 2 h _ { 2 } = 3 h _ { 3 }$

A balloon starts ascending in a vertical upward direction at a constant rate of

$5m/{s^2}$ . After

$2s$ of its flight begins, a ball has been dropped from it, find the maximum height attained by ball. Also, calculate the position of the balloon relative to ground when ball strikes the ground. (Take g =

$10m/{s^2}$ )

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$10m,56m$
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$5m,137.1m$
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$15m,56m$
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$15m,25m$

A body starts from rest and travels with a uniform acceleration of

$3 ms^{-2}$ and then decelerates at a uniform rate of

$2 ms^{-2}$ again to come to at rest, total time of travel is

$10$ sec. Find the maximum velocity during the journey and the total displacement

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$12 ms^{-1}, 60 m$
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$60 ms^{-1}, 6 m$
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$24 ms^{-1}, 60 m$
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$60 ms^{-1}, 24 m$

An elevator ascends with an upward acceleration of

$0.2$

${ ms }^{ -2 }$ . At the instant its upward speed is

$2m/s$ , a loose bolt

$5$ m high from the floor drops from the ceiling of the elevator. The time taken by the bolt to strike the floor and the distance it has fallen are

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$1s, 1.9 m$
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$1s, 2.9 m$
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$1s, 4.9 m$
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$1s, 3.9 m$

A particle starts from rest and experiences constant acceleration for 6 seconds. If it travels a distance

$d _ { 1 }$ in the first two seconds, a distances

$d _ { 2 }$ in the next two seconds and a distances

$d _ { 3}$ in the last two seconds, then

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$d _ { 1 } : d _ { 2 } : d _ { 3 } = 1 : 1 : 1$
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$d _ { 1 } : d _ { 2 } : d _ { 3 } = 1 : 2 : 3$
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$d _ { 1 } : d _ { 2 } : d _ { 3 } = 1 : 3 : 5$
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$d _ { 2 } : d _ { 2 } : d _ { 3 } = 1 : 5 : 9$

Two cars A and B are moving with same speed of 45 km/h along same direction. If a third car C coming from the opposite direction with a speed of 36 km/h meets two cars in an interval of 5 min, the distance of seperation of two A and B should be :

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6.75 km
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7.25 km
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5.55 km
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8.35 km

A car, starting from position of rest, moves with constant acceleration

$8\ m/s^{2}$ . Then it moves with constant deceleration

$y$ and becomes stationary. If it aquires maximum velocity is

$10\ m/s$ then total time takes by it is

$20\ sec$ , then

$y=........$

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$(15/8)\ m/s^{2}$
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$(3/4)\ m/s^{2}$
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$(8/15)\ m/s^{2}$
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$(5/18)\ m/s^{2}$

A man weighing 80 kg is standing at the centre of a flat boat and he is 20m from the shore. He walks 6m on the boat towards the the shore and then halts. The boat weight 200 kg. How far is he from the shore at the end of this time?

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11.2 m
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13.8 m
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14.3 m
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15.4 m

A man runs at a speed of

$4\mathrm { ms } ^ { - 1 }$ to overtake a standing bus. When he is

$6\mathrm { m }$ behind the door

$\mathrm { t } = 0 ,$ the bus starts to move forward and continues with a constant acceleration of

$1.2\mathrm { ms } ^ { - 2 } .$ If the man reaches the door in time

$t$ , then the can be obtained by the equation

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$4 t = 6 + 0.6 t ^ { 2 }$
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$1.2 t ^ { 2 } = 4 t$
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$4 t ^ { 2 } = 1.2 t$
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$6 + 4 t = 0.6 t ^ { 2 }$

A rickshaw covers a journey of 200 m in 17 s. It starts from rest with constant acceleration of

$2 m/s^2$ , then moves some distance with constant speed and finally decelerates at

$5 m/s^2$ until it stops at the end of journey. Its maximum velocity is

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10 m/s
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20 m/s
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30 m/s
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40 m/s

A sprinter has to cover a total run of

$100 m$ . She increases her speed from rest under an uniform acceleration of

$1.0 m/s^{2}$ up to three quarters of the total run and covers the last quarters with uniform speed. The time she takes to cover the first half, and to cover the second half of the run will be

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$9s , 3.25 s$
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$10s, 4.28 s$
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$10s, 5.25 s$
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$6.25 s, 10s$

For the given graph, correct relation is:-

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$a_1=a_2=a_3$
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$$a_1
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$a_1>a_2>a_3$
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$a_1>a_3>a_2$

A partical is observed to be at rest as seen from a frame of reference. We can conclude that -

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the frame is inertial
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Resultant force on the particle is zero.
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Resultant froce on the particle is zero , if the frame is inertial
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There is a non-zero resultant force ,if the frame is non- inertial.

A ball rolls off the top of stairway with a horizontal velocity of magnitude

$1.8{ m } / { s }.$ The states are

$0.20 { m }$ high and

$0.2 { m }$ wide. Which step will the ball hit first ?

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First
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Second
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Third
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Fourth

A freely falling body crosses points A and B with velocities V and 2V respectively. The velocity of that body at C such that AB = BC will be

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$\sqrt{3}$ V
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4 V
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$\sqrt{5}$ V
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$\sqrt{7}$ V

Which of the following equation represents the motion of a body moving with constant finite acceleration? In these equation, y denotes the displacement in time t and p.q. and r are arbitary constant ;

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$y = {(p+q)}^2{(r + pt)}$
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y = p + tqr
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$y = {(p+t)}{(q+t)}{(r+1)}$
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$y = {(p+qt)}r$

Two cars

$A$ and

$B$ are moving with same speed of

$45$

$\mathrm { km } / \mathrm { hr }$ along same direction. If a third car C coming from opposite direction with a speed of

$36$

$\mathrm { km } / \mathrm { hr }$ meets two cars in an interval of

$5$ minutes, the distance of separation of two cars

$A$ and B should be

$( in\:km )$

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6.75
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7.25
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5.55
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8.35

A man throws ball with the same speed vertically upwards one after the other at an interval of 2 seconds. What should be the speed of the throw so that more than two balls are in the sky at any time ?

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Atleast 0.8m/s
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Any speed less than 19.6 m/s
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Only with speed 19.6m/s
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More than 19.6m/s

An object is thrown vertically upward and experiences an air resistance opposing its motion with magnitude proportional to its speed. Which of the following graphs best represents the variation of the acceleration, a, of the object with time t, starting from the moment when the object leaves one's hand up to the time when it returns to the ground? (Take downward as positive)

If a smooth stright tunnel is cut at any orientation through earth, then a ball released from one end will reach the other end in time (neglect earth rotation)

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$84.6$ minutes
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$42.3$ minutes
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$8$ minutes
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depends on orientation

A student is standing at

$50 \ m$ from a bus, and at the same time when bus starts with a constant acceleration of

$1 \ m /{s }^{ 2 }$ , student runs behind it to catch it, at a uniform speed

$u.$ Find minimum value of

$u$ so that student can catch the bus:

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5 m/s
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8 m/s
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10 m/s
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12 m/s

A man m=80 kg is standing ona trolley of mass 320 kg on a smooth surface. If man starts walking on trolley along rails at a speed of 1

$ms^{-1}$ , then after 4 sec, his displacement relative to ground is :-

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4 m
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4.8 m
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3.2 m
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6 m

A car accelerates from rest at a constant rate

$\alpha$ for some time after which it decelerates at a constant rate

$\beta$ to come to rest. If the total then, the time elapsed it

$t$ , the distance travelled by the car is

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$\dfrac{1}{2} \left(\dfrac{\alpha \beta}{\alpha + \beta}\right)t^2$
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$\dfrac{1}{2} \left(\dfrac{\alpha+ \beta}{\alpha \beta}\right)t^2$
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$\dfrac{1}{2} \left(\dfrac{\alpha^2+ \beta^2}{\alpha \beta}\right)t^2$
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$\dfrac{1}{2} \left(\dfrac{\alpha^2- \beta^2}{\alpha \beta}\right)t^2$

A man weighing

$80 kg$ and a lady weighing

$40 kg$ standing on a lift. The lift moves downwards with uniform acceleration of

$7.35m/{ s }^{ 2 }$ . The factors by which their weights are reduced apparently will be in the ratio of

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$2 : 1$
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$1 :1$
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$4 : 1$
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$6 :5$

The speed

$v$ of a particle moving along a straight line, when it is at distance

$( \mathrm { x } )$ from a fixed point of the line is given by

$v ^ { 2 } = 108 - 9 x ^ { 2 } .$ (all quantities are in cgs units)

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the motion is uniformly accelerated along a straight line
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the magnitude of the acceleration at a distance 3 $\mathrm { cm }$ from the point is $27$ $\mathrm { cm } / \mathrm { sec } ^ { 2 } .$
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the motion is simple harmonic about the given fixed point.
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the maximum displacement from the fixed point is 4 $\mathrm { cm }$

After some time the bus will be left behind. If the bus continues moving with the same acceleration, after what time from the beginning, the bus will overtake the cyclist?

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10 s
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12 s
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14 s
0%
16 s

Explanation

$t_1$ sec was taken, when both bus and cycle meets.

distance traveled by bus = distance traveled by cycle - 96 = s

$s = ut + \dfrac{1}{2}at^2 \\ s = 0*t + \dfrac{1}{2} * 2 * t^2$

$s = t^2$

time taken by cycle =

$\dfrac{s+96}{20}$

$t = \dfrac{t^2+96}{20} \\$

$20t = t^2 + 96$

$t^2 -20t + 96 = 0$

$(t-8)(t-12) = 0$

t = 8 sec and t = 12 sec

so first time at 8 sec and second time at 12 sec both bus and cycle will meet.

Hence option 'B' is correct.

A stone dropped from the top of a towe travels 25m in the last second of its motion g=10m

${ s }^{ -2 }$ height of the tower is

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45m
0%
90m
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72m
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135m

A ball is released from the top of a tower of height h metre. it takes T second to reach the ground, what is the position of the ball in T/3?

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h/9 m from the ground
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7h/9 m from the ground
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8h/9 from the ground
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17h/18 m from the ground

The acceleration time graph of a particle moving along a straight line is as shown in the figure. After what time the particle acquires

$0$ velocity.

Note :- The particle initially starts at rest.

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$12 \ s$
0%
$8 \ s$
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$5 \ s$
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$16 \ s$

The motion of a body falling from rest in a resisting medium is described by the equation

$\dfrac{dv}{dt}=a-bv$ , where a and b are constant. The velocity at any time t is?

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$a(1-b^{2t})$
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$\dfrac{a}{b}(1-e^{-bt})$
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$abe^{-t}$
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$ab^2(1-t)$

A ship moves at 40 km/h due north and suddenly moves towards east through

$9\mathring { 0 }$ and continuous to move with the same speed. Then the change in its velocity is

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$zero$
0%
$40\sqrt { 2 }$ km/h north-east
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$40\sqrt { 2 }$ km/h south-east
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$None$

Wind is blowing at a harbor with a speed of 72 km/hr and the flag on the mast of a boat anchored at harbor fluters a long the north -east direction. If the boat starts moving at a speed of

$51 km/hr$ due north, the direction of the flag is (approximately).

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Towards east
0%
towards west
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towards north
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towards south

what would be the approximate retardation to be given by jet pack along for safe landing?

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$5g\ ms^{-2}$
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$2g\ ms^{-2}$
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$4g\ ms^{-2}$
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$Cannot\ be\ determined$

A train starts from station with an acceleration

$1 m/s^2$ . A boy who is

$48$ m behind the train with a constant velocity

$10$ m/s, the minimum time after which the boy will catch the train is?

0%
$4.8$ sec
0%
$8$ sec
0%
$10$ sec
0%
$12$ sec

The velocity-time plot for a particle moving on a straight line is shown in the figure, then?

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The particle has a constant acceleration
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The particle has never turned around
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The average speed in the interval $0$ to $10$ s in the same as the average speed in the interval $10$ s to $20$ s
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Both (a) and (c) are correct

Explanation

Option A is true because in a velocity-time graph, the slope (which shows the rate of change of velocity) denotes acceleration. In this graph the slope is constant, so acceleration is constant.

Option B is not true because at

$t=10 s$ the velocity changes its sign from positive to negative, which means it has turned around.

Option C is true because in

$0$ to

$10 s$ speed changes from

$10$ to

$0 \,m/s$ and in

$10$ to

$20 s$ speed changes from

$0$ to

$10 \,m/s$ . So average speed will be the same.

Two particles start from the same point along the same straight line. The first moves with constant velocity '

$v$ ' and the second with constant acceleration '

$a$ '. During the time that elapses before the second catch the first, the greatest distance between the particles is:

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$\dfrac{v^{2}}{a}$
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$\dfrac{v^{2}}{2a}$
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$\dfrac{2v^{2}}{a}$
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$\dfrac{v^{2}}{4a}$

The shortest distance between the motorcyclist and the car is

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20 m
0%
10 m
0%
40 m
0%
30 m

A sailor drops a wrench from the top of a sailboats vertical mast while the boat is moving rapidly and steadily straight forward. Where will the wrench hit the deck?

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ahead of the base of the mast
0%
at the base of the mast
0%
behind the base of the mast
0%
on the windward side of the base of the mast
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None of the choices (a) through (d) is true.

CBSE Questions for Class 11 Engineering Physics Motion In A Straight Line Quiz 15 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Motion In A Straight Line Quiz 15 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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