MCQ Questions for Class 11 Engineering Physics Motion In A Straight Line Quiz 13

Two trains one of length

$100\ m$ and another of length

$125m$ , are moving in mutually opposite directions along parallel lines, meet each other, each with speed

$10m/s$ . If their acceleration are

$0.3m/{s}^{2}$ and

$0.2m/{s}^{2}$ respectively, then the time they take to pass each other will be

0%
$5\ s$
0%
$10\ s$
0%
$15\ s$
0%
$20\ s$

Explanation

$\textbf{Step 1: Drawing initial and final position [Refer Fig. 1 and 2]}$

$\textbf{Step 2: Solving in frame of reference of Train 2.}$

Initial speed

$u_{{1}/{2}} = 10 - (-10)m/s$

$= 20\ m/s$

Acceleration

$a_{\frac {1}{2}} = 0.3 - (-0.2) m/s^{2}$

$= 0.5\ m/s^{2}$

Displacement

$s = 100 - (-125) m$

$= 225\ m$

$\textbf{Step 3: Apply equation of motion}$

Since acceleration is constant, therefore applying equation of motion

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

$\Rightarrow$

$225 = 20t + \dfrac {1}{2} (0.5) t^{2}$

$\Rightarrow$

$t^{2} = 80 t - 900 = 0$

Solving the quadratic eqn

${t = 10s}$

Hence option

$B$ is correct.

2112094_1033134_ans_4ec1aa2cb9e04b8abb382a27a77ebde2.png

Two guns are mounted (fixed) on two vertical cliffs that are very high from the ground as shown in figure. The muzzle velocity of the shell from

${G}{1}$ is

${u}{1}$ and that from

${G}{2}$ is

${u}{2}$ . The guns aim exactly towards each other The ratio

${u}{1}:{u}{2}$ such that the shells collide with each other in air is (Assume that there is no resistance of air)

0%
$1:2$
0%
$1:4$
0%
$1:6$
0%
will not collide for any ratio

A man driving his car with constant acceleration on a highway. He crosses two check post with speed

$30\sqrt { 3 }\ m/s$ and

$10\sqrt { 3 }\ m/s$ . The speed of car when he is exactly

${1/3}^{rd}$ of total distance from pole

$I$ is

0%
$35\sqrt { 2 }\ m/s$
0%
$50\ m/s$
0%
$50\sqrt { 2 }\ m/s$
0%
$10\sqrt { 19 }\ m/s$

A car starting from rest, is accelerated at a constant rate

$\alpha$ until it attains a speed

$V$ . It is then retarded at a constant rate

$\beta$ until it comes to rest. The average speed of the car during its entire journey is :

0%
Zero
0%
$\dfrac{\alpha v }{2}$
0%
$\dfrac {\beta v }{2}$
0%
$\dfrac{v }{2}$

Explanation

The average speed is given as,

${v_{avg}} = \dfrac{s}{t}$

${v_{avg}} = \dfrac{{\dfrac{{{V^2}}}{{2\alpha }} + \dfrac{{{V^2}}}{{2\beta }}}}{{\dfrac{V}{\alpha } + \dfrac{V}{\beta }}}$

${v_{avg}} = \dfrac{V}{2}$

A particle starts from rest and travels a total distance of 18m along a straight path.The first half of the distance was travelled with a uniform acceleration of 2

$ms^{-2}$ and the rest uniform velocity.The average velocity for the whole journey is (in

$ms^{-1}$ )

0%
3
0%
4
0%
6
0%
9

A barometer kept in an elevator reads

$76 cm$ when it is at rest. What will be barometric reading when the elevator accelerates upwards?

0%
$\dfrac {(g+a)h_0}{g}$
0%
$\dfrac {gh_0}{g+a}$
0%
$\dfrac {a}{g} h_0$
0%
$\dfrac{g}{a} h_0$

Explanation

When a body moves upward the acceleration due to gravity is,

${g'} = g + a$

Hence, the pressure is

$P = h_0\rho \left( {g + a} \right)$

$= \dfrac{{h_0\left( {g + a} \right)}}{g}$

Hence, the barometer reading will be more than the measured value at rest.

A machine gun is mounted on a

$2000 kg$ vehicle on a horizontal smooth road (friction negligible). The gun fires

$10$ bullets per sec with a velocity of

$500 m/s$ . If the mass of each bullet be

$10 g$ , what is the acceleration produced in the vehicle?

0%
$25 cm/s^2$
0%
$0.025 m/s^2$
0%
$0.50 cm/s^2$
0%
$50 m/s^2$

Explanation

The momentum of each bullet is given as,

${P_b} = {m_b}{v_b}$

$= 10 \times {10^{ - 3}} \times 500$

$= 5\;{\rm{kgm/s}}$

The force on the vehicle will be equal to change in momentum in $1\;{\rm{sec}}$ .

The force is given as,

$F = n{P_b}$

$= 10 \times 5$

$= 50\;{\rm{N}}$

The acceleration of the vehicle is given as,

$a = \dfrac{F}{{{m_v}}}$

$= \dfrac{{50}}{{2000}}$

$= 0.025\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}$

Thus, the acceleration produced in the vehicle is $0.025\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}$ .

In the system shown in figure

$m_A = 4m, m_B = 3 m$ and

$m_C = 8 m$ . Friction is absent everywhere. String is light and in-extensible. If the system is released from rest find the acceleration of block B

0%
$\dfrac{g}{8}$ (leftward)
0%
$\dfrac{g}{2}$ (leftward)
0%
$\dfrac{g}{6}$ (rightward)
0%
$\dfrac{g}{4}$ (rightward)

Explanation

When the system is released from the position of rest, the equations of motions are given as,

$4a = 4g - T$ (1)

$T = 3a$ (2)

From the equation (1) and (2) it can be written as,

$7a = 4g$

$a = \dfrac{{4g}}{7}$

$a \approx \dfrac{g}{2}$

Thus, the acceleration of block B is $\dfrac{g}{2}$ towards the left side.

A particle initially at rest is subjected to two forces. One is constant, the other is a retarding force proportion at to the particle velocity. In the subsequent motion of the particle.

0%
The acceleration will increase from zero to a constant value.
0%
The acceleration will decrease from its initial value to a zero.
0%
The velocity will increase from zero to a maximum & then decrease.
0%
The velocity will increase from zero to a constant value.

Explanation

Let,

Constant force $=K_1$

Variable force $=-K_2V$

Net Force $F=K_1-K_2V$

Acceleration $a=\dfrac{K_1-K_2V}{m}$

As the velocity increases the net force will decrease. When both retarding and constant forces become equal then net force is zero. Hence the velocity is maximum and acceleration will increase from zero to constant.

A body is projected vertically up at t = 0 with a velocity of 98 m/s. Another body is projected from the same point with same velocity after 4 seconds. Both bodies will meet at t =

0%
6 s
0%
8 s
0%
10 s
0%
12 s

Explanation

Both the bodies have been thrown with same initial velocity

So,

$h_1 = h_2$

Time of flight form for 1st body = t seccond

and for 2nd body = (t- 4) second (because it is projected 4sec later)

SO,

$h_1 = h_2$

$ut - \dfrac{1}{2} gt^2 = 4(t - 4) - \dfrac{1}{2}g (T - 4)^2$

By solving the equation

$t = 12 second$

Hence (D) option is correct.

$30\ kg$ block rests on a rough horizontal surface. A force of

$200\ N$ is applied on the block. The block acquires a speed of

$4\ m/s$ , starting from rest in

$2\ s$ . What is the value of coefficient of friction?

0%
${10}/{3}$
0%
${\sqrt 3}/{10}$
0%
$0.47$
0%
$0.185$

Explanation

The acceleration is given as

$v = u + at$

$a = 2\;m/{s^2}$

The force applied on the horizontal surface is such that it just overcome the static friction

$F - \mu N = ma$

$200 - \mu \cdot 30 \cdot 10 = 30 \times 2$

The static friction is $0.47$

A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20 s. Find its acceleration. Find the force acting on it if its mass is 7 tonnes (Hint: I tonne 1000 kg.)

0%
$N=85\times 10^{4}N$
0%
$N=5\times 10^{4}N$
0%
$N=6.85\times 10^{8}N$
0%
$N=6.85\times 10^{4}N$

Explanation

Here,

$s=400m$

and

$t=20sec$ .

By relation

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

$s=400=0+\dfrac{1}{2}\times a\times 20\times 20$

$a=2m/s^{2}$

also,

$g\sin \theta=2$

$\sin \theta=\dfrac{1}{5}$

$\therefore \cos \theta=\dfrac{\sqrt{25-1}}{5}=\dfrac{2\sqrt{6}}{5}$

and,

$N=mg\cos \theta$

$=7\times 10^{4}\times \dfrac{2\sqrt{6}}{5}$

$N=6.85\times 10^{4}N$

The driver of a train. A travelling at a speed of

$60 \ km/h$ applies brakes and brakes and retards the train uniformly. The train stops in

$5 \ s$ , another train

$B$ is travelling on the parallel track with a speed of

$36 \ kmph$ . This driver also applied the brakes and the train retards uniformly. The train

$B$ stops in

$10 \ s$ then distance travelled by each train after the brakes were applied is:

0%
$d_A = 41.6 m, d_B = 50 m$
0%
$d_A = 42.6 m, d_B = 20 m$
0%
$d_B = 50 m, d_A = 41.6 m$
0%
$d_B = 80 m, d_A = 50 m$

A body starting from rest is moving with a uniform acceleration

$5m/s^2$ fro time

$10s$ and after that with uniform acceleration

$10m/s^2$ for time

$15s$ then

0%
Average acceleration of the body is $7.5 m/s^2$
0%
Average acceleration of the body is $8.0 m/s^2$
0%
total distance travelled by body is $1875m$
0%
total distance travelled by body is $2125m$

The times taken by a block wood (initially at rest ) to slide down a smooth inclined plane

$9.8m$ long (angle of inclination is

$30^o$ ) is

0%
$\dfrac{1}{2}s$
0%
$1s$
0%
$2s$
0%
$4s$

A lift initial at rest top floor of a building moves downwards with a constant speed of

$5/ms$ for first

$8s$ . Therefore the support ropes are cut . The time taken by the lift to reach the ground floor from the instant its ropes were cut is. [take

$g=10m/s^2$ and height of building

$=100m$ ]

0%
$4s$
0%
$3s$
0%
$2s$
0%
$5s$

A particle starts from rest, accelerates at

$2\ ms^{-2}$ for

$10\ s$ and then goes for constant speed for

$30\ s$ and then deceleration at

$4\ ms^{-2}$ till stops after next. What is the distance travelled by it?

0%
$750\ m$
0%
$800\ m$
0%
$700\ m$
0%
$850\ m$

Explanation

Displacement in first, $10\,\sec$

${{s}_{1}}=\dfrac{1}{2}{{a}_{1}}{{t}_{1}}^{2}=\dfrac{1}{2}\times 2\times {{10}^{2}}=100\,m$

Speed achieve in first $10\,\sec$

${{v}_{1}}={{u}_{1}}+{{a}_{1}}{{t}_{1}}=0+2\times 10=20\,m/s$

Displacement in $30\,\sec$ , acceleration is zero.

${{s}_{2}}={{v}_{1}}{{t}_{2}}=20\times 30=600\,m$

Displacement in last due to deacceleration, ${{a}_{3}}=-4\,m/{{s}^{2}}$

${{v}^{2}}-{{v}_{1}}^{2}=2as$

${{s}_{3}}=\dfrac{0-{{20}^{2}}}{2\times (-4)}=50\,m$

Net displacement

${{S}_{total}}={{s}_{1}}+{{s}_{2}}+{{s}_{3}}=100+600+50=750\,m$

Total displacement is $750\,m$

A block rests on a truck moving with a velocity of

$2m/s$ . The coefficient of friction between box and truck is 0.1 . The driver applies the brake, and the truck starts to decelerate uniformly and stops in

$s$ . The total distance travelled by the box w.r.t the truck is

0%
$0.75\ m$
0%
$0.375\ m$
0%
$1. 5\ m$
0%
$2\ m$

An object performs upwards journey under gravity in time

$t$ . The downwards journey is performed in time

0%
$t$
0%
$t^2$
0%
$2t$
0%
$t/2$

A wooden plank of mass 20kg is resting on a smooth horizontal floor. A man of mass 60kg starts moving from one end of the plank to the other end. The length of the plank is 10m. Find the displacement of the plank over the floor when the man reaches the other end of the plank.

0%
5m
0%
6.5m
0%
2.5m
0%
7.5m

A lift whose cage is

$3\ m$ high is moving up with an acceleration of

$2\ m/s^{2}$ . A piece of stone is dropped from the top of the cage of the lift when its velocity is

$8\ m/s$ . If

$g = 10\ m/s^{2}$ , then the stone will reach the floor of the lift after.

0%
$0.7\ s$
0%
$0.5\ s$
0%
$0.4\ s$
0%
$0.3\ s$

Explanation

The time taken to reach the stone on the floor is given as,

$s = ut + \frac{1}{2}\left( {a + g} \right){t^2}$

$3 = 8 \times t + \frac{1}{2}\left( {2 + 10} \right){t^2}$

$3 = 8 \times t + 6{t^2}$

$t = 0.3052\;\sec$

A horse is running with constant acceleration

$\dfrac {g}{\sqrt {3}}$ . A small ball is projected by a horce rider at an angle

$\theta$ with horizontal with respect to house.The value of

$\theta$ such that the ball again caught by the boy

0%
$36^{o}$
0%
$45^{o}$
0%
$60^{o}$
0%
$90^{o}$

A trolley was moving horizontally on a smooth ground with velocity

$v$ with respect to the earth. Suddenly a man starts running from rear end of the trolley with a velocity

$\dfrac {3v}{2}$ with respect to the trolley in opposite direction. If the length of the trolley is

$L$ , find the displacement of the man with respect to earth when he reaches the starting point on the trolley. [Mass if the trolley is equal to the mass of the man].

0%
$\dfrac {L}{3}$
0%
$\dfrac {2L}{3}$
0%
$\dfrac {4L}{3}$
0%
$zero$

water drop are falling from calling at regular interval of time in such away that sixth drop if going to be deleched when

$1^{st}$ drop it just hitting the ground. If time interval between two consecitive drope is

$0.5$ second. Then the height of celling it.

0%
$31.25m$
0%
$45m$
0%
$62.5m$
0%
$20m$

A force of

$50$

$N$ is required to push a car on a level road with constant speed of

$10$

${m/s}$ . The mass of the car is

$500$

$kg$ . What force should be applied to make the car accelerate at

$1$

${m/s^2}$ ?

0%
$550$ $N$
0%
$450$ $N$
0%
$500$ $N$
0%
$2500$ $N$

Explanation

Since $50\;{\rm{N}}$ force is required for the car to move with constant velocity.

Additional force applied to move a car of $500\;{\rm{kg}}$ with acceleration of $1\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}$ is given by

$F = ma$

$F = 500 \times 1$

$F = 500\;{\rm{N}}$

Thus total force applied

${F_T} = 500 + 50$

$F = 550\;{\rm{N}}$

Determine in time in which the smaller block reaches other end of bigger block in the figure.

0%
$4\ s$
0%
$8$
0%
$2.19\ s$
0%
$2.13\ s$

Explanation

Both block will not move together because friction force between block is not sufficient to accelerate 8Kg block with same acceleration

Drawing

$FBD$ of

$2kg$ block

$f_r=$ frictional force

$f_r=\mu Mg$

for vertical balance,

$M=20\ N$

$fr=0.3\times 20=6\ N$

Now let the block moves with acceleration

$a_{1}$

$\Rightarrow f-fr=ma_{1}$

$10-6=2\times a_{1}$

$\Rightarrow a_{1}=2\ m/\sec^{2}$

Drawing

$FBD$ of block of

$8\ kg$ mass

$\Rightarrow$ Let

$a_{2}$ be its acceleration

$a_2=\dfrac{f_r}{m_2}=\dfrac{.3\times 20}{8}$

$a_{2}=\dfrac{6}{8}=\dfrac{3}{4}\ m/\sec^{2}$

Relative acceleration of both the blocks

$a=a_{1}-a_{2}=2-\dfrac{3}{4}=\dfrac{5}{4}$

distance to be covered

$=3\ m$ initial velocity

$=0$

$S=ut+\dfrac{1}{2}at^{2}$

$=3=0+\dfrac{1}{2}\times \dfrac{5}{4}\times t^{2}$

$\Rightarrow t=2.19\ \sec$

Option

$C$

1456877_1080178_ans_7e95e1ef674b4c49b1d19a7770f8e2ab.png

A body moving with uniform acceleration along a straight line covers

$21\ m$ in the fifth seconds of its motion and

$41\ m$ in the tenth seconds of its motion. What is its initial velocity?

0%
$2\ m/s$
0%
$3\ m/s$
0%
$4\ m/s$
0%
$5\ m/s$

A car is travelling on a straight road. The maximum velocity the car can attain is

$24\ ms^{-1}$ . The maximum acceleration and deceleration it can attain are

$1\ ms^{-2}$ and

$4\ ms^{-2}$ respectively. The shortest time the car takes to start from rest and come to rest in a distance of

$200$ meter is

0%
$22.4\ second$
0%
$30.0\ second$
0%
$11.2\ second$
0%
$5.6\ second$

A particle starts from rest travels a distance

$x$ with uniform acceleration, then it travels a distance

$2x$ with uniform speed, finally, it travels a distance

$3x$ with uniform retardation and comes to rest. If the whole motion of the particle is a straight line, then the ratio of its average velocity to maximum velocity is

0%
$6:7$
0%
$4:5$
0%
$2:3$
0%
$3:5$

A ball is thrown vertically upwards with a velocity of

$10 \, ms^{-1}$ . It returns to the ground with a velocity of

$9 ms^{-1}$ .If

$g=9.8 ms^{-2}$ , then the maximum height attained by the ball is nearly (assume air resistance to be uniform)

0%
$5.1 m$
0%
$4.1 m$
0%
$4.61 m$
0%
$5 m$

Explanation

Given,

$u=10m/s$

$v_m=0m/s$

$g=-9.8m/s^2$

From 3rd equation of motion,

$2gs=v^2-u^2$

$-2\times 9.8\times s=0^2-10^2$

$s=\dfrac{10\times 10}{2\times 9.8}$

$s=5m$

The correct option is D.

A train starts from station

$A$ with uniform acceleration

$\alpha$ for some distance and then goes with uniform retardation

$\beta$ for some more distance to come to rest at station

$B$ . The distance between station

$A$ and

$B$ is

$4\ km$ and the train takes

$4$ minutes to complete this journey. If

$\alpha$ and

$\beta$ are in km

$(min)^{-2}$ then

$\dfrac {1}{\alpha} + \dfrac {1}{\beta} =$ .

0%
$2$
0%
$4$
0%
$\dfrac {1}{2}$
0%
$\dfrac {1}{4}$

A car starts from rest and moves in a straight line with a constant acceleration

$\alpha$ . After time

$t_0$ , brakes are applied which causes retardation of magnitude

$\beta$ and car finally stops. The distance travelled by the car is

0%
$\alpha t_0^2 \cfrac{(\alpha + \beta)}{4\beta}$
0%
$\alpha t_0^2 \cfrac{(\alpha + \beta)}{2\beta}$
0%
$\beta t_0^2 \cfrac{(\alpha + \beta)}{2\alpha}$
0%
$\beta t_0^2 \cfrac{(\alpha + \beta)}{4\alpha}$

Explanation

Distance traveled in

$t_0$ seconds

$S_1=0\times t_0+\dfrac{1}{2}\alpha t^2$ (u=0 which is at rest)

Final velocity after

$t_0$ seconds,

$V_s=0+\alpha t_0$

Final velocity after applying retardation

$\beta=0$

Initial velocity =final velocity after

$t_0$ seconds

$0=(V_s)^2-2\beta(s_2)$ (

$s_2$ is the distance traveled after

$\beta$ retardation)

$s_2=\dfrac{(V_s)^2}{2\beta}$

$s_2=\dfrac{\alpha^2t_0^2}{2\beta}$

$s_1+s_2=\dfrac{1}{2}\alpha t_0^2(\alpha+\dfrac{\alpha}{\beta})$

$=\dfrac{\alpha t_0^2(\alpha+\beta)}{2 \beta}$

A person sees a stone fall from the top of a distant building and notes that it takes 1.5 s for the stone to fall the last third of the way to the ground. With ignoring air resistance, the height of the cliff in meters is

0%
100m
0%
200km
0%
300m
0%
none

A car starts from rest and moves on straight line with constant acceleration

$a_{0}$ . After time

$t_{0}$ , brakes is applied which cause of deceleration of magnitude as initial acceleration. The distance travelled by car is

0%
$\dfrac {1}{2}a_{0}t_{0}^{2}$
0%
$2a_{0}t_{0}^{2}$
0%
$a_{0}t_{0}^{2}$
0%
$\dfrac {2}{3}a_{0}t_{0}^{2}$

A particle starts from rest and traverses a distance

$l$ with uniform acceleration, then moves uniformly over a further distance

$3\ l$ under uniform retardation. Assuming entire motion to be rectilinear motion the ratio of average speed over the journey to the maximum speed on its way is

0%
$\dfrac {1}{5}$
0%
$\dfrac {2}{5}$
0%
$\dfrac {3}{5}$
0%
$\dfrac {4}{5}$

Two bodies were thrown simultaneously from the same point at time

$t=0$ , one straight up and other at an angle

${30^ \circ }$ with horizontal. the initial speed of the each body is equal to

$30m/s$ . The magnitude of relative displacement between the two bodies at

$t=2s$ will be

$(g = 10m/{s^2})$

0%
$15m$
0%
$45m$
0%
$60m$
0%
$30m$

The speed-time graph of a particle moving along a fixed direction is shown in figure. Then which of the following are correct?

0%
distance travelled by particle in $10$ sec is $60m$
0%
average speed in $10$ sec in $6m/s$
0%
average speed between $t=2,t=6$ sec is $9m/s$
0%
average speed between $t=2,t=6$ sec is $6m/s$

From a motorboat moving downstream with a velocity

$2\ m/s$ with respect to river, a stone is thrown. The stone falls on an ordinary boat at the instant when the motorboat collides with the ordinary boat. The velocity of the ordinary boat with respect to the river is equal to zero. The river flow velocity is given to be

$1\ m/s$ . The initial velocity vector of the stone with respect to earth is

0%
$2\hat {i} + 20\hat {j}$
0%
$3\hat {i} + 40\hat {j}$
0%
$3\hat {i} + 50\hat {j}$
0%
$2\hat {i} + 50\hat {j}$

Explanation

Ref. image I

$\dfrac{d}{t} = \dfrac{20}{t} = 2$

$\therefore t = 10 sec$

Ref. image II

$S = ut + \dfrac{1}{2} at^2$

$0 = u_y t + \dfrac{1}{2} (-g) t^2$

$\dfrac{1}{2} gt^2 = u_y t$

$\therefore u_y = \dfrac{1}{2} \times 10 \times 10 = 50 m/s$

Ref. image III

$u_B - u_r = 0$

$u_B = u_r$

$u_B = 1 m/s$

w.rt. river

$2m/s$ V

$\rightarrow$

$4x = 3 m/s = 2m/s + u_B$

$V_{store} = 3 \hat{i} + 50 \hat{j}$

Hence, option C is correct

1456747_1115861_ans_77f3447eeda9408c9e102d391671eedd.png

A stone is allowed to fall from top of a building and covers half of the height in last second of its motion. The time taken by stone to reach the bottom of building is?

0%
$(2-\sqrt{2})s$
0%
$(2+\sqrt{7})s$
0%
$2s$
0%
$1s$

Explanation

Let

$‘h’$ be the height of the tower,

$‘n’$ be the time taken to reach the ground

$(h,n,>0)$ .

Using the formula

$S$ =

${{ut} + {{\dfrac{1}{2}}{at^2}}}$

$h$ =

$0$ +

${{\dfrac{1}{2}}{(9.8)n^2}}$ …………………….(1)

According to the problem, the stone travels

${\dfrac{h}{2}}$ in the

$n^{th}$ second.

So, by using

$S_n$ =

$u$ +

${{\dfrac{a}{2}}(2n - 1)}$ we get

${\dfrac{h}{2}}$ =

$0$ +

${{\dfrac{9.8}{2}}(2n - 1)}$ [as initial velocity is

$0$ ]

Using (1),

${\dfrac{{\dfrac{1}{2}(9.8)n^2}}{2}}$ =

${{\dfrac{9.8}{2}}(2n - 1)}$

${\dfrac{n^2}{2}}$ =

$2n – 1$

$n$ =

$(2 + √2)$ s or

$(2 - √2)$ s

A man

$m=80kg$ is standing on a trolley of mass

$320kg$ on a smooth surface. If man starts walking on trolley along rails at a speed of

$1m{s}^{-1}$ , then after

$4$ sec, his displacement relative to ground is

0%
$4m$
0%
$4,8m$
0%
$3.2m$
0%
$6m$

A trolley is moving horizontally with a constant velocity of v m/s w.r.t. earth. A man starts running from one end of the trolley with a velocity

$1.5v m/s$ w.r.t. to trolley. After reaching the opposite end, the man return back to starting end and continues running with a velocity of 1.5 v m/s w.r.t. the trolley in the backward direction. If the length of the trolley is L then the displacement of the man with respect to earth during the process will be :

0%
$2.5 L$
0%
$1.5 L$
0%
$\dfrac { 5L }{ 3 }$
0%
$\dfrac { 4L }{ 3 }$

Which of the following equations represents the motion of a body moving with constant finite acceleration? in these equations,

$y$ denote the displacement in time

$t$ and

$p, q$ and

$r$ are arbitary constants:

0%
$y = (p + qt)^2(r + pt)$
0%
$y = p + tqr$
0%
$y = (p + t)(q + t)(r + 1)$
0%
$y = (p + qt)r$

A body is projected vertically upwards with speed

$40 m/s$ . The distance traveled by the body in the last second of upward journey is ( take g=

$9.8 m/s^2$ and neglect effect of air resistance)

0%
4.9 m
0%
9.8 m
0%
12.4 m
0%
19.6 m

Two cars are moving in the same directions with the same speed of

$30$ km/hr. They are separated by

$5$ km. What is the speed of the car moving in the opposite direction if it meets the second cars at an interval of

$4$ minute?

0%
$45$ km/hr
0%
$60$ km/hr
0%
$105$ km/hr
0%
None of these

A particle starts from rest with constant acceleration for 20 s. If it travels a distance

$y_1$ in the first 10 s and a distance

$y_2$ in the next 10 s then

0%
$y_2=2y_1$
0%
$y_2=3y_1$
0%
$y_2=4y_1$
0%
$y_2=5y_1$

A particle moves in a straight line with a constant acceleration. it changes its velocity from

$10\ ms^{-1}$ to

$20\ ms^{-1}$ while passing through a distance

$135\ m$ in

$t$ seconds. The value of

$t$ is

0%
$9$
0%
$10$
0%
$1.8$
0%
$12$

Explanation

Given,

$u=10m/s$

$v=20m/s$

$S=135m$

$t=?$

From 3rd equation of motion,

$2as=v^2-u^2$

$a=\dfrac{v^2-u^2}{2s}$

$a=\dfrac{20\times 20-10\times 10}{2\times 135}$

$a=1.11m/s^2$

From 1st equation of motion,

$v=u+at$

$20=10+1.11t$

$10=1.11t$

$t=\dfrac{10}{1.11}$

$t=9sec$

The correct option is A.

A body is allowed to slide from the top along a smooth inclined plane of length 5m at an angle of inclination

$30^o$ . If

$g = 10ms^{-2}$ , time taken by the body to reach the bottom of the plane is

0%
$\dfrac{\sqrt{3}}{2}s$
0%
$1.41s$
0%
$\dfrac{1}{\sqrt{2}}s$
0%
$2s$

A ball is thrown vertically upwards with a velocity of

$10 ms^-1$ . IT returns to the ground with a velocity of

$9 ms^-1$ . If

$g = 9.8 ms^-2$ , then the maximum height attained by the ball is nearly ( assume a resistance to be uniform)

0%
5.1 m
0%
4.1 m
0%
4.61 m
0%
5 m

Explanation

Given,

$u=10m/s$

$g=-9.8m/s^2$

$v=0m/s$

Lets consider,

$H=$ maximum height attained by the ball

From 2nd equation of motion,

$2as=v^2-u^2$

$H=\dfrac{v^2-u^2}{2g}$

$H=\dfrac{0^2-10\times 10}{-2\times 9.8}$

$H=5.1m$

The correct option is A.

A parachutist after bailing out falls for

$10s$ without friction. When the parachute opens he descends with an acceleration of

$2\ m/s^{2}$ against his direction and reached the ground with

$4\ m/s$ . From what height he has dropped himself?

$(g=10m/s^{2})$

0%
$500m$
0%
$2496m$
0%
$2996m$
0%
$4296m$

A stone A is thrown at an angle of

$45^0$ to the horizontal. Another stone B is thrown with

$20ms^{-1}$ horizontally as shown in figure. They collide in mid-air. Find the distance BC

0%
$144$ m
0%
$48$ m
0%
$72$ m
0%
$108$ m

CBSE Questions for Class 11 Engineering Physics Motion In A Straight Line Quiz 13 - 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 13 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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