A. 19.6 m/s.

B. any speed less than 19.6 m/s.

C. more than 19.6 m/s.

D. 9.8 m/s.

**Right Answer is: **

#### SOLUTION

**Q. 302** What is the technique used for measuring the large time intervals.

B. any speed less than 19.6 m/s.

C. more than 19.6 m/s.

D. 9.8 m/s.

RADAR stands for radio detection and ranging. Radar is a device to detect the position of aero plane in the sky. Radar works on the principle of reflection of waves called echo. Radio waves are transmitted in all direction and when they hit the target or aero plane, the waves are reflected back. The reflected waves are received by antenna of receiver. Time gap between the transmission and reception of waves is noted and then by using the formula S = V(T/2) distance of aero plane can be measured.

A. 19.6 m/s.

B. any speed less than 19.6 m/s.

C. more than 19.6 m/s.

D. 9.8 m/s.

**Right Answer is: **

#### SOLUTION

**Q. 303** What type of method is most suitable to measure the time

B. any speed less than 19.6 m/s.

C. more than 19.6 m/s.

D. 9.8 m/s.

Very large intervals of time are measured by the method of carbon dating or radioactive dating. Carbon dating is used to measure the age of fossils and uranium dating is used to find the age of rock.The age can be calculated by noticing the ratio of number of radioactive atoms, which are decayed to the no. of atoms undecayed.

A. 19.6 m/s.

B. any speed less than 19.6 m/s.

C. more than 19.6 m/s.

D. 9.8 m/s.

**Right Answer is: **

#### SOLUTION

**Q. 304** What are errors? Explain two types of errors?

B. any speed less than 19.6 m/s.

C. more than 19.6 m/s.

D. 9.8 m/s.

The phenomenon which repeats itself after fixed interval of time is used to measure the time. E.g. the rotation of earth about its own axis. Heartbeat can be used to measure the time but above all the cesium atom clock is most accurate.

A. 19.6 m/s.

B. any speed less than 19.6 m/s.

C. more than 19.6 m/s.

D. 9.8 m/s.

**Right Answer is: **

#### SOLUTION

**Q. 305** Convert 25 Joule into erg.

B. any speed less than 19.6 m/s.

C. more than 19.6 m/s.

D. 9.8 m/s.

The difference in the true and observed value of any physical quantity is called errors.e.g the velocity of sound in air is 332 m/s. After experiment the value of velocity is found to be 331.3m/s so the difference in the two is called error. The two types of errors are random error and gross error. Random error occurs by chance, its cause is not known. It can be minimized by taking large number of readings and then mean of them. Gross error occurs due to the carelessness of the observer.

A. 19.6 m/s.

B. any speed less than 19.6 m/s.

C. more than 19.6 m/s.

D. 9.8 m/s.

**Right Answer is: **

#### SOLUTION

**Q. 306** Derive the formulae for velocity (v) of water waves that may depend upon their wavelength λ, density of water ∂ and acceleration due to gravity g.

B. any speed less than 19.6 m/s.

C. more than 19.6 m/s.

D. 9.8 m/s.

By using dimensional formulae we can convert 25 Joule into erg. Joule is the unit of energy of dimensional formulae ML^{2}T^{-2}

A. 19.6 m/s.

B. any speed less than 19.6 m/s.

C. more than 19.6 m/s.

D. 9.8 m/s.

**Right Answer is: **

#### SOLUTION

**Q. 307** If x = a + bt + ct^{2} where x is in metres and t in seconds, write the units of a, b, c.

B. any speed less than 19.6 m/s.

C. more than 19.6 m/s.

D. 9.8 m/s.

A. 19.6 m/s.

B. any speed less than 19.6 m/s.

C. more than 19.6 m/s.

D. 9.8 m/s.

**Right Answer is: **

#### SOLUTION

**Q. 308** What do you mean by significant figures? Discuss their rule also.

B. any speed less than 19.6 m/s.

C. more than 19.6 m/s.

D. 9.8 m/s.

a + bt + ct^{2} should have unit of distance i.e. metre.

Unit of a = m

Unit of bt = m

Unit of b= m/t= m/s

Unit of ct^{2} = m

Unit of c = m/s^{2}

A. 19.6 m/s.

B. any speed less than 19.6 m/s.

C. more than 19.6 m/s.

D. 9.8 m/s.

**Right Answer is: **

#### SOLUTION

**Q. 309** A body A moves with a uniform acceleration ‘a’ and zero initial velocity. Another body B starting from the same point moves in the same direction with a constant velocity v. The two bodies will meet after a time

B. any speed less than 19.6 m/s.

C. more than 19.6 m/s.

D. 9.8 m/s.

The significant figures are those figures, which are considered in the measurement. The following are the rule to measure the significant figures:

1: All non zero digits are significant e.g. the digit 3476, there are 4 significant figures.

2: All the zeros between non zero digits are significant e.g. 14005, there are 5 significant figures.

3: All zeros on the right of the last non-zero digit in the decimal part are significant e.g. 0.00800 has 3 significant figures.

4: All zeros on the right of non-zero digit are NOT significant e.g. 378000 has 3 significant figures.

5: In a number less than one, all zeros to the right of decimal point and to the left of non-zero digit are not significant. E.g. 0.0084 has 2 significant.

A. 3a/2v.

B. v/2a.

C. v/3a.

D. 2v/a.

If the two bodies meet after time t, then the

Distance traveled by body A in time t, S_{A} = 0 + (1/2)at^{2}

Distance traveled by body B in time t, S_{B} = vt
Now, S_{A} = S_{B} (1/2) at^{2 }= vt
So, t = 2v/a

A. v_{0 }- b/2 + c

B. v_{0 }+ b/2 + 3c

C. v_{0 }+ b/2 + c/3

D. v_{0 }+ b + c

A. 100 m/s

B. 150 m/s

C. 250 m/s

D. 450 m/s

A. S_{2 }= S_{1}

B. S_{2 }= 4S_{1}

C. S_{2 }= 2S_{1}

D. S_{2 }= 3S_{1}

Distance covered in first 10 seconds,

Distance covered in 20 seconds,

So, the distance covered in second 10 seconds, S_{2 }= 200a - 50a = 150a

So, S_{2 }= 3S_{1}

A.
(h_{1}/h_{2})^{1/2}

B.
(m_{1}/m_{2})^{1/2}

C.
(h_{2}/h_{1})^{1/2}

D.
(m_{2}/m_{1})^{1/2}

A. 42.5 km/hr.

B. 40.0 km/hr.

C. 38.2 km/hr.

D. 37.5 km/hr.

Time taken to cover first half of the distance = Time taken to cover next half of the distance = Total time taken to cover the whole distance = So, average speed =

A. 0.5 m/s^{2}.

B. 1.09 m/s^{2}.

C. 3.8 m/s^{2}.

D. 4.9 m/s^{2}.

Remaining mass = (1000 - 100)m = 900 m

Force = mg = 100g (as removal of 100 kg mass gives force to the balloon)

A. parabola.

B. hyperbola.

C. straight line.

D. ellipse.

For upward motion, v = u – gt

So, velocity-time (v-t) graph is a straight line with slope (-g) and an intercept u on velocity axis.

A. 180 km/h

B. 120 km/h

C. 80 km/h

D. 60 km/h

A. 15 m

B. 25 m

C. 35 m

D. 50 m

A. 5 m

B. 6 m

C. 8 m

D. 10 m

A. 26/9.

B. 4/3.

C. 7/5.

D. 2.

A. the force is zero.

B. there is a force, which opposes motion.

C. there is a force towards motion.

D. there is only gravitational motion.

Body is being retarded at B, so force opposes the motion.

A. 2.

B. (2)^{1/2}

C. 1/2.

D. 1/(2)^{1/2}

Since the elevator is moving with uniform speed so the acceleration of the coin is g. So, the time taken by the coin to reach the floor is,

A. 2.2 m/s and 0.4 m/s.

B. 2.2 m/s and 1.8 m/s.

C. 4 m/s and 0.4 m/s.

D. 5 m/s and 0.8 m/s.

Let the speeds of the two particles be u and v.

Then u +v=4 and u-v=0.4

From the above two equations we get u=2.2 m/s and v=1.8 m/s

A. vertical and constant.

B. vertical and variable.

C. horizontal and constant.

D. horizontal and variable.

Since, the two particles cover equal vertical distances, their vertical velocities are equal and so there is no relative motion in vertical direction. The horizontal velocity of the particle thrown in vertical direction is zero but the horizontal velocity of the other is not zero. So, there is only a relative velocity in horizontal direction, which remains constant in magnitude.

A. 60 seconds.

B. 50 seconds.

C. 40 seconds.

D. 30 seconds.

Relative speed of first train with respect to second train = (30 - 20) = 10 m/s The additional distance traveled by the first train to pass the second train = (300 + 200) = 500.

A. .

B. .

C. .

D. .

A. velocity 10 m/s, speed 10 m/s.

B. velocity zero, speed zero.

C. velocity zero, speed 10 m/s.

D. velocity 10 m/s, speed zero .

Distance traveled in one lap = 2 r = 2 3.14 100

Time for one lap = 62.8 sec

So, average speed =

Displacement = 0 and so velocity = 0

A. 2a.

B. 2.

C. a/2.

D. 1/2.

Distance traveled =2a, Displacement=a2.

A. always = 1.

B. always 1.

C. may be 1.

D. always 1.

Displacement is the shortest distance travelled by a body between the initial point (say A) and the final point (say B). For example

A. zero.

B. 2r.

C. /2r.

D. /r.

Displacement is the shortest distance travelled by a body between the initial point and the final point. Here, net displacement is zero

A. velocity is zero.

B. time taken is zero.

C. velocity increases.

D. velocity decreases.

Displacement is the shortest distance travelled by a body between the initial point (say A) and the final point (say B) and it remains constant always between A and B. However, the distance can change.

Velocity = dx/dt = zero.

A. acceleration.

B. distance.

C. displacement.

D. average velocity.

A. 200 m.

B. 300 m.

C. 500 m.

D. 1250 m.

Let first man catch the second man after t second

70t=50(t+10)

or, t =25s

Distance covered by first man in 25 s =50x25=1250 m

A. 36.11 m/s.

B. 34.67 m/s.

C. 30 m / s.

D. 25.45 m/s.

V_{BA }= V_{B}+ V_{A} as both are coming opposite directions.

A. parallel to time axis.

B. parallel to velocity axis.

C. diagonal to time axis.

D. diagonal to velocity axis.

In the uniform motion, the body moves with constant velocity.

Hence, the rate of change of velocity with time is zero i.e. slope of velocity-time graph is zero which corresponds to a straight line parallel to the time axis.

The average velocity of the particle is

A. 3 ms^{-1}.

B. 4 ms^{-1}.

C. 5 ms^{-1}.

D. 6 ms^{-1}.

A. non-uniform acceleration.

B. uniform acceleration.

C. uniform retardation.

D. zero acceleration.

Acceleration = dv/dt = 3 x 2t = 6t

As it depends upon time t; hence, acceleration is non- uniform.

A. .

B. .

C.

D. 1 : 2.

Let ‘r’ be the radius of the semicircular path.

Circumference = 2r

‘l’ is the semicircular distance travelled.

Diameter is shortest distance travelled between two points in a semicircle.

Magnitude of displacement = Diameter = (2l)/

Distance: Displacement =

A. 35 m.

B. 25 m.

C. 12.5 m.

D. 10 m.

From the 2^{nd} equation of kinematics

s = ut + ½ at^{2}

for a vertical downward motion of the body, we have

45 = 0 + ½ x 10 x t^{2} or t = 3s

From

D_{3} = u + ½ a (2x 3-1) = 0 + ½ x 10 x 5 = 25 m.

A. 90 m.

B. 100 m.

C. 115 m.

D. 150 m.

Given, u =, a = 20 m/s^{2}, D_{8} =?

From

D_{n} = u + ½ a (2n-1)

D_{8} = 0 + ½ x 20 x [(2x8)-1]

= 10 x 15 = 150m.

A. s_{1}= s_{2}.

B. s_{2 }= 2s_{1}.

C. s_{2 }= 3s_{1}.

D. s_{2 }= 4s_{1}.

A. 4 hour.

B. 5 hour.

C. 6 hour.

D. 8 hour.

Distance covered by A = (20 x t) km

Distance covered by B = (10 x t) km

Total distance = (20 t + 10 t) =150 km

t = 5 hour

A. 45 m/s^{2}.

B. 40 m/s^{2}.

C. 35 m/s^{2}.

D. 30 m/s^{2}.

Velocity, v = (dx/dt) = 10+30 t,

acceleration, a = dv/dt = d^{2}x / dt^{2}= 30 m/s^{2}.

A.

B.

C.

D.

Slope (dv/dt) = -ve value

But dv/dt = acceleration.

A. 20 m.

B. 45 m.

C. 80 m.

D. 150 m.

A. 7.25.

B. 6.75.

C. 5.55.

D. 4.75.

Relative velocity of car C w.r.t. car A or B

= 45 – (-36) = 81 km/h

Distance between car A and car B = 81 x (5/60)

= 6.75 km.

A. 2.

B. 3.

C. 4.

D. 8.

Given, s = 6t^{2}-t^{3
}velocity,

v = ds/dt = 12t – 3t^{2}

or v = 12t – 3t^{2}.......... (1)

If velocity is zero, then equ(1) becomes

0 = 12t – 3t^{2}

Either t =0 s or, t= 4.0s
Since time can never be zero; so, t = 4.0 s.

A. 4 m/s.

B. 8m/s.

C. 16 m/s.

D. 4 m/s.

Average speed = (total distance)/(total time) = 200/25=8 m/s

A. zero.

B. 2 m/s.

C. 4 m/s.

D. 8 m/s.

In a circular, the shortest distance travelled in a half revolution is diameter.

A. 98 m.

B. 66 m.

C. 44.1 m.

D. 39.2 m.

For a falling body,

h = ( 1/2 )gt^{2}
^{ }h = ( 1/2 ) x 9.8 x 9 = 44.1 m

A. 1:n.

B. .

C. .

D. .

A. 16 s.

B. 17 s.

C. 18 s.

D. 19 s.

Here s = 0, v= 80 ms^{-1}, a=10 m/s^{2}, t=?

From the 2^{nd} equation of kinematics

s = ut + ½ at^{2}

we have,

0 = 80t + ½ x (-10) x t^{2}

or, t = 16 s.

A. the lighter body.

B. the heavier body.

C. both simultaneously at the same point.

D. both simultaneously at the different points.

Due to more mass, the heavier body will experience more gravitation pull. So, it will be able to overcome the opposing force of air friction easily than the lighter body.

A. 4 ms^{-2}.

B. 2 ms^{-2}.

C. -1 ms^{-2}.

D. -2 ms^{-2}.

A. have zero height.

B. have acceleration.

C. be colourless.

D. be dimensionless.

In motion under gravity, at highest point of its motion velocity is zero but acceleration ‘a’ = ‘g’ in downward direction.

A. 2v.

B. v.

C. v/2.

D. 0.

A. 5.

B. 7.

C. 9.

D. 11.

Let s = thickness of the plank

a = uniform retardation of the plank
u = initial velocity

The velocity of the bullet after passing through a plank is

So, the number of planks required to stop the bullet completely = 1 +10 = 11.

A. 19 m/s.

B. 29 m/s.

C. 39 m/s.

D. 49 m/s.

The distance covered by the particle will be same in the 5^{th} and 6^{th} second if it reaches at the highest position in the 5^{th} seconds. Using t = 5s, a = -9.8 m/s^{2}, v=0 and u=?

Now, from the 1^{st} equation of kinematics

v = u + at

A. 50 m.

B. 55 m.

C. 60 m.

D. 65 m.

A. non-uniform motion.

B. uniform motion.

C. variable speed.

D. variable speed.

During uniform motion, a body covers or travels equal displacements in equal intervals of time. Hence, the average and instantaneous velocities have same value as the uniform velocity.

A. 45 ms^{-1}.

B. 55 ms^{-1}.

C. 90 ms^{-1}.

D. 110 ms^{-1}.

Maximum velocity

= area under acceleration time graph

= ½ x 12 x 15

= 90 ms^{-1}.

A. uniform acceleration.

B. uniform velocity.

C. non-uniform acceleration.

D. zero velocity.

A. D.

B.

A.

C. B.

D. C.

Instantaneous velocity = slope of distance – time graph at the given instant of time.

As slope is highest at C, so instantaneous velocity

= ds/dt = max at C.

A. 8 m.

B. 10 m.

C. 80 m.

D. 90 m.

As the time of ascent (t_{1}) = time of descent (t_{2}); therefore, as per question,

t_{1} + t_{2 }= 8

or, t_{2} + t_{2} = 8s

or, t_{2} = 4s
Taking vertical downward motion of the ball from the highest point to the surface of the earth, we have

u =0; a = 10 m/s^{2}; t= 4s, S=?

From the 2^{nd} equation of kinematics,

S = ut + ½ at^{2
}or, S = 0 + ½ x 10 x 4^{2}

= 80m.

A. a dimensionless quantity.

B. time independent.

C. a vector quantity.

D. a scalar quantity.

Acceleration is defined as the ratio of change in velocity and the corresponding time taken by the object. So, acceleration has magnitude as well as direction like velocity.

A. once.

B. twice.

C. thrice.

D. quadrupled.

Acceleration is the rate of change of velocity. Velocity is the rate of displacement. Hence, time occurs twice in its unit, i.e., metre per second per second.

A. 4.9 m.

B. 3.9 m.

C. 2.9 m.

D. zero.

From the 2^{nd} equation of kinematics

s = ut + ½ at^{2}

acceleration, a=g = 9.8 m/s^{2}

Distance covered,

s = ½ gt^{2} = ½ x 9.8 x (1)^{2} = 4.9 m.

A. 1 m/s.

B. 2 m/s.

C. 3 m/s.

D. 5 m/s.

u= 0; s = 15 m; t= 15s; a=? ; v=?

From the 2^{nd} equation of kinematics

s = ut + ½ at^{2}

or, 15 = 0 + (½ x a x 15^{2})

or, a = (2/15) m/s^{2}

From the 1^{st} equation of kinematics

v = u + at

or, v = 0 + [2/15] x 15= 2 m/s.

A. 1 m/s.

B. 2 m/s.

C. 3 m/s.

D. 5 m/s.

**Right Answer is: **

#### SOLUTION

**Q. 370** A jet plane is moving with a velocity of 800 km/hr. The gases are ejecting from rear of jet plane with velocity of 1600 km/hr with respect to jet. Find the velocity of gases w.r.t a person on ground.

B. 2 m/s.

C. 3 m/s.

D. 5 m/s.

(1)The graph shows that body has uniform velocity throughout the motion.

(2)This graph shows that body has some initial velocity and moving with uniform acceleration.

(3)This graph shows that body is moving with some velocity and moving with retardation.

A. 1 m/s.

B. 2 m/s.

C. 3 m/s.

D. 5 m/s.

**Right Answer is: **

#### SOLUTION

_{gp}= 800km/hr(-i)+1600km/hr(i)

**Q. 371** A particle is moving along X axis the position is given by
x = K + jt^{2}
where K = 8m and j = 4 m/s^{2} t is time. Find velocity of particle at t = 0, t= 3 sec.

B. 2 m/s.

C. 3 m/s.

D. 5 m/s.

Consider the directions as according to origin. +ve towards +X axis and negative towards –ve X axis. Jet plane and gases moves in opposite direction. Velocity of jet plane Vg = 60 km/hr with respect to +x-axis

V_{gp} = V_{gj} – V_{pj} ( V_{gj}=vel of gas w.r.t to jet, V_{jp}=vel of jet w.r.t person )

V_{gp}= V_{gj}+_{ } V_{jp}

= 1600-800

= 800km/hr

So, this will be velocity of gases w.r.t person standing on ground.

A. 1 m/s.

B. 2 m/s.

C. 3 m/s.

D. 5 m/s.

**Right Answer is: **

#### SOLUTION

**Q. 372** Draw the position time graphs for two objects initially occupying different positions but having zero relative velocity.

B. 2 m/s.

C. 3 m/s.

D. 5 m/s.

X = k + Jt^{2 }

K = 8m, j = 4 m/s^{2 }

X = 8 + 4t^{2}

To get velocity differenciate both sides v = = (8 + 4t^{2})

v 4 x 2 t = 8t

so = v = velocity

when t = 0, v = 0

when t = 3 sec. v = 8 X 3

= 24 m/s.

A. 1 m/s.

B. 2 m/s.

C. 3 m/s.

D. 5 m/s.

**Right Answer is: **

#### SOLUTION

**Q. 373**

B. 2 m/s.

C. 3 m/s.

D. 5 m/s.

The position time graphs for two objects initially occupying different positions but having zero relative velocity are parallel to each other.

A. 1 m/s.

B. 2 m/s.

C. 3 m/s.

D. 5 m/s.

**Right Answer is: **

#### SOLUTION

**Q. 374** Two trains A and B of length 600m each are moving on two parallel tracks with a uniform speed of 60km/hr. in same direction with A ahead of B. B accelerates with 1m/s^{2 }to overtake

A. If after 50sec. the guard of B just brushes past the driver of A, what was the original distance between them?

B. 2 m/s.

C. 3 m/s.

D. 5 m/s.

A. Body is at rest.

B. Body is moving with constant velocity.

C. Body is moving with constant velocity coming back to origin

D. The velocity of body is increasing

E. The body is moving with decreasing velocity.

F. Body first moves with increasing velocity and then velocity

decreases.

A. If after 50sec. the guard of B just brushes past the driver of A, what was the original distance between them?

u_{A} = 60 km/hr = 60 ( 5/18 ) m/s = ( 50/3 ) m/s

S_{1} = ( 50/3 ) 50 = ( 2500/3 ) m

u_{B} = 60 km/hr = 60 ( 5/18 ) m/s = ( 50/3 ) m/s

a = 1 m/ s^{2} t = 50Sec.

S = ut + ( 1/2 ) at ^{2}

^{ } =^{ }( 50/3 ) 50 + ( 1/2 ) 1 50 50

= ( 6250/3 ) m.

The original distance was

( 6250/3 ) - ( 2500/3 ) = ( 3750/3 ) = 1250 m

A. 1 m/s.

B. 2 m/s.

C. 3 m/s.

D. 5 m/s.

**Right Answer is: **

#### SOLUTION

**Q. 376** Define relative velocity. Explain its concept in detail.

B. 2 m/s.

C. 3 m/s.

D. 5 m/s.

A. 1 m/s.

B. 2 m/s.

C. 3 m/s.

D. 5 m/s.

**Right Answer is: **

#### SOLUTION

**Q. 377** Derive equation of motion v = u + at, by graphical method.

B. 2 m/s.

C. 3 m/s.

D. 5 m/s.

Relative velocity of an object A with respect to another object B is the comparison of their velocities. It can also be understood by time rate at which object A changes its positions w.r.t B. Consider two objects A and B moving with velocities V_{1} and V_{2}

Displacement of A at t = 0, x_{01}

Displacement of B at t = 0, x_{02}

If X_{1} and X _{2} is the displacement of A and B after time t.

displacement of A in time t

( X_{1} – X_{01}) = V_{1} T - (1)

displacement of B in time t

( X_{2} – X_{02}) = V2 t - (2)

( X_{2} – X_{1}) = (X_{02} + X_{10}) +(V_{2} – V_{1}) t

Let X_{2} – X_{1} = X and X_{0 }= (X_{02} + X_{01})

X = X_{0} + (V_{2} – V_{1}) t

Relative velocity of B w.r.t A is

V_{BA} = V_{2} – V_{1}

When the two velocities are equal V_{2} = V_{1}_{
}V_{2} – V_{1} = 0

X = X_{0}.

The two bodies will remain at constant distance from each other. This can be shown by the graph (1)

When two bodies move with unequal velocity the graph will be (2)

In general , we can write

V_{ab}= V_{a}- V_{b}

If they are moving with respect to a third quantity, then

V_{ab}= V_{ac}- V_{bc}

A. 1 m/s.

B. 2 m/s.

C. 3 m/s.

D. 5 m/s.

**Right Answer is: **

#### SOLUTION

Let ‘u’ be the initial velocity of particle at t = 0, ‘v’ is final velocity after time ‘t .

‘‘A’ represents the velocity of body at time ,t = 0;

‘B’ represents the velocity of body at time, t = t

OA = CD = u

BC = v - u,

OD = t.

acceleration = slope of graph

v - u = at,

v = u + at

**Q. 378** A body is dropped from the top of a tower, which falls through 40m during the last two seconds of its fall. What is the height of tower ?(g= 10 m/s^{2})

B. 2 m/s.

C. 3 m/s.

D. 5 m/s.

Consider an object moving along a st. line with uniform acceleration ‘a’.

Let ‘u’ be the initial velocity of the object at time , t=0 and ‘v’ be the final velocity of the object after time ‘t’.

Let ‘u’ be the initial velocity of the object at time , t=0 and ‘v’ be the final velocity of the object after time ‘t’.

Below is the velocity-time graph of the motion which is st . line.

Let ‘u’ be the initial velocity of particle at t = 0, ‘v’ is final velocity after time ‘t .

‘‘A’ represents the velocity of body at time ,t = 0;

‘B’ represents the velocity of body at time, t = t

OA = CD = u

BC = v - u,

OD = t.

acceleration = slope of graph

v - u = at,

v = u + at

A. 1 m/s.

B. 2 m/s.

C. 3 m/s.

D. 5 m/s.

**Right Answer is: **

#### SOLUTION

**Q. 379** Derive distance velocity relation by graphical method.

B. 2 m/s.

C. 3 m/s.

D. 5 m/s.

A. 1 m/s.

B. 2 m/s.

C. 3 m/s.

D. 5 m/s.

**Right Answer is: **

#### SOLUTION

**Q. 380** Derive the equations of motion v = u + at, s= ut + (1/2)at^{2}, v^{2} – u^{2} = 2as by calculus.

B. 2 m/s.

C. 3 m/s.

D. 5 m/s.

Consider an object moving along a st. line with uniform acceleration ‘a’.

Let ‘u’ be the initial velocity of the object at time , t=0 and ‘v’ be the final velocity of the object after time ‘t’.

‘S’ be the distance travelled by the object in time t.

Let ‘u’ be the initial velocity of the object at time , t=0 and ‘v’ be the final velocity of the object after time ‘t’.

‘S’ be the distance travelled by the object in time t.

Below is the velocity-time graph of the motion which is st . line.

In figure,

OA = CM = u.

BM = OE = v.

BC = v-u,

BM = OE = v.

BC = v-u,

OM = AC = t

Distance traversed in time interval t, is

S = Area of trapezium OABM

A. 1 m/s.

B. 2 m/s.

C. 3 m/s.

D. 5 m/s.

**Right Answer is: **

#### SOLUTION

**Q. 381** Two balls are thrown simultaneously one vertically upwards with speed of 20m/s from the ground and other vertically downwards from height of 40m with same speed and in same line of motion. Determine the point where they will collide.

B. 2 m/s.

C. 3 m/s.

D. 5 m/s.

Consider an object moving along a st. line with uniform acceleration ‘a’.

Let ‘u’ be the initial velocity of the object at time , t=0

and ‘v’ be the final velocity of the object after time ‘t’.

Let ‘u’ be the initial velocity of the object at time , t=0

and ‘v’ be the final velocity of the object after time ‘t’.

Acceleration,

a = (dv/dt)

or, dv = a.dt

(1) Velocity-time relation:

Let at instant time, t, dv be the change in velocity in time interval dt.

Integrating , within the condition of motion

(i.e. time from 0 to t and velocity from u to v ), we get,

(2) Distance-time relation :

Let at an instant t, dx be the displacement of the object in time interval dt.

Instantaneous velocity ‘v’ is given by

(3)Velocity-displacement relation:

A. 1 m/s.

B. 2 m/s.

C. 3 m/s.

D. 5 m/s.

**Right Answer is: **

#### SOLUTION

**Q. 382** Read the assertion and reason carefully to mark the correct option out of the options given below.

Assertion: A particle is moving in a circular path in a clockwise direction. Then angular velocity will be perpendicular to the linear velocity and angular acceleration.

Reason: The angular velocity will act in a downward direction perpendicular to the plane of rotation.

B. 2 m/s.

C. 3 m/s.

D. 5 m/s.

Suppose two balls meet at height 'X' above ground after time t sec. from start.

For upward motion,

u = 20m/s, g= - 9.8m/s^{2
}From 2nd equation of kinematics,

s = ut + 1/2at^{2} ^{ }

X = 20t - 1/2(9.8 t^{2})

or,X = 20t – 4.9 t^{2}...........(1)^{ }

For downward motion:

40 - X = 20t + 4.9 t^{2}..........(2)

Adding equ (1) & (2) , we get,

40t = 40,

t = 1 sec.

From equ(1)

X = 20 – 4.9

or,X = 15.1m.

Both balls collide after 1 sec at height 15.1 above the ground.

Assertion: A particle is moving in a circular path in a clockwise direction. Then angular velocity will be perpendicular to the linear velocity and angular acceleration.

Reason: The angular velocity will act in a downward direction perpendicular to the plane of rotation.

A. Assertion is true but reason is false.

B. Assertion and reason both are false.

C. Both assertion and reason are true and the reason is the correct explanation of the assertion.

D. Both assertion and reason are true but reason is not the correct explanation of the assertion.

According to right-hand thumb rule if fingers point in the direction of rotation then thumb represents the angular velocity vector. Here, the particle is moving in the clockwise direction so angular velocity will act in a downward direction perpendicular to the rotational plane. As in circular motion centripetal acceleration acts towards centre and perpendicular to velocity vector acting in the tangential direction. Both velocity and acceleration are acting in the plane of rotation. So, angular velocity will be perpendicular to both linear velocities as well as centripetal acceleration.

Assertion: When dot product and cross product of any two vectors is zero, then one of the vectors must be null vector.

Reason: A null vector has a zero magnitude.

A. Assertion is true but reason is false.

B. Assertion and reason both are false.

C. Both assertion and reason are true and the reason is the correct explanation of the assertion.

D. Both assertion and reason are true but reason is not the correct explanation of the assertion.

\begin{array}{l}\text{Let us assume two vectors}\overrightarrow{\text{a}}\text{and}\overrightarrow{\text{b}}\text{and \theta is the angle between them}\text{. Given,}\\ \text{dot product of the vectors is zero, i}\text{.e}\overrightarrow{\text{a}}\text{.}\overrightarrow{\text{b}}\text{=}\left|\text{a}\right|\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left|\text{b}\right|\text{cos \theta \xb0= 0 and cross product}\\ \text{of vectors is zero, i}\text{.e}\text{.}\text{\hspace{0.17em}}\overrightarrow{\text{a}}\text{\xd7}\overrightarrow{\text{b}}\text{=}\left|\text{a}\right|\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left|\text{b}\right|\text{sin \theta \xb0 = 0}\text{. In this case, if}\left|\overrightarrow{\text{a}}\right|\text{and}\left|\overrightarrow{\text{b}}\right|\text{are}\\ \text{not zero, then both}\mathrm{sin}\text{\theta and}\mathrm{cos}\text{\theta must be simultaneously zero}\text{.But is not}\\ \text{possible so one of the vectors}\text{\hspace{0.17em}}\text{must be null vector}\text{.}\end{array}

Assertion: A river is flowing towards the east at a speed of 4 ms

Reason: He will cross the river in the shortest time when he swims along the direction of resultant velocity.

A. Assertion is true but reason is false.

B. Assertion and reason both are false.

C. Both assertion and reason are true and the reason is the correct explanation of the assertion.

D. Both assertion and reason are true but reason is not the correct explanation of the assertion.

\begin{array}{l}\begin{array}{l}\text{Suppose width of the river is w and swimmer starts swimming with velocity}\\ {\text{v}}_{\text{s}}\text{making an angle of \theta}\xb0\text{with the vertical direction}\text{.Then effective velocity}\\ {\text{of the swimmer is v}}_{\text{s}}\text{cos \theta}\text{.}\end{array}\hfill \\ \text{Time taken to cross the river, t =}\frac{\text{W}}{{\text{v}}_{\text{s}}\text{cos}\text{\hspace{0.17em}}\text{\theta}}\hfill \\ \begin{array}{l}\text{For time to be minimum, cos \theta should have maximum value, i}\text{.e}\text{. \theta = 0\xb0}\text{.}\\ \text{The swimmer should swim due north}\text{.}\end{array}\hfill \end{array}

Assertion: A man holds the water hose at 45° so that he can wet the more area of the lawn.

Reason: The water stream will get the greatest range when launched at an angle of 45°.

A. Assertion is true but reason is false.

B. Assertion and reason both are false.

C. Both assertion and reason are true and the reason is the correct explanation of the assertion.

D. Both assertion and reason are true but reason is not the correct explanation of the assertion.

In the absence of air resistance, the only force acting on the water stream is the force due to gravity, due to which the water stream follows the parabolic path. The water stream will wet the maximum area when it reaches the maximum distance or range. The range in projectile motion is given by the formula, \begin{array}{l}\text{R=}\text{\hspace{0.17em}}\frac{{\text{v}}^{\text{2}}\text{sin}\text{\hspace{0.17em}}\text{2\theta}}{\text{g}}\\ \text{Given, angle \theta = 45\xb0}\text{.}\text{\hspace{0.17em}}\\ \therefore {\text{R}}_{\text{max}}\text{=}\text{\hspace{0.17em}}\frac{{\text{v}}^{\text{2}}\text{\hspace{0.17em}}\text{sin}\text{\hspace{0.17em}}\text{2}\left({\text{45}}^{\text{\xb0}}\right)}{\text{g}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\frac{{\text{v}}^{\text{2}}}{\text{g}}\end{array} Thus, the water stream will reach maximum area when launched at 45°.

Assertion: The speedometer needle represents the radial acceleration of the body.

Reason: The radial acceleration is a measure of the rate of change in the magnitude of the velocity vector, whereas the tangential acceleration is a measure of the rate of change of the direction of the velocity vector.

A. Assertion is true but reason is false.

B. Assertion and reason both are false.

C. Both assertion and reason are true and the reason is the correct explanation of the assertion.

D. Both assertion and reason are true but reason is not the correct explanation of the assertion.

The tangential acceleration is the measure of the rate of change in the magnitude of the velocity vector, thus, always coincides with the speed, and directed along with the tangent drawn at any point on trajectory. The normal acceleration is a measure of the rate of change of the direction of the velocity vector and is directed perpendicular to the tangential acceleration in the direction of the concavity of the curve.

Assertion: When a ball is dropped from the train moving at a constant speed, for a person standing on the ground, the path of the ball is parabola however the motion appears straight line to the person on the train.

Reason: The path of the ball is a straight line as seen from the moving train because both ball and train are moving at the same speed. But, with respect to ground, the path of the ball is parabola as the vertical velocity of the ball changes with time.

A. Assertion is true but reason is false.

B. Assertion and reason both are false.

C. Both assertion and reason are true and the reason is the correct explanation of the assertion.

D. Both assertion and reason are true but reason is not the correct explanation of the assertion.

The path of the ball depends upon the position of the observer. For an observer inside the train, the path of the ball is vertically downward, as the horizontal velocity of the ball and the train are the same. But, when this is observed from the ground, the path of the ball is a parabola because the vertical velocity of the ball is influenced by the acceleration due to gravity.

Assertion: The resultant of three vectors cannot be zero if they are coplanar.

Reason: The three unequal vectors should form a polygon to give zero resultant vector.

A. Assertion is true but reason is false.

B.

Assertion and reason both are false.

C. Both assertion and reason are true and the reason is the correct explanation of the assertion.

D. Both assertion and reason are true but reason is not the correct explanation of the assertion.

For getting a zero-resultant vector, the three vectors can be arranged in a polygon. If three vectors are coplanar then their resultant vector can be zero, that is when the resultant of two of them is equal in magnitude and opposite to the direction of the third vector.

A. 8 m/s, 1 m/s.

B. 6 m/s, 4 m/s.

C. 8 m/s, 4 m/s.

D. 4 m/s, 6 m/s.

Let u and v be the speeds of the two cars

u+v=10,

When the cars approach each other

u-v=6

2u=16

or u=8 m/s

2v=10-8=2 m/s or v= 1 m/s.

A. 10 s

B. 15 s

C. 20 s

D. 25 s

Let bike catch the car after t second
60t=40(t+10)

or, t=20 s

A. 40 s.

B. 50 s.

C. 90 s.

D. 100 s.

Relative velocity of one train w.r.t. second = 50-32 =18 km/h = 5 ms^{-1}
Total distance traveled = 200 + 250 =450 m,
Time taken, t = 450/5 = 90 s.

A. 4 s.

B. 6 s.

C. 8 s.

D. 9 s.

Relative velocity of one train w.r.t. second = 40+20 =60 km/h = 50/3 ms^{-1}
Total distance traveled = 100 + 50 =150 m,
Time taken, t = ( 150 x 3) / 50 = 9 s.

A. 15 km/h

B. 12 km/h

C. 10 km/h

D. 5 km/h

Let t_{1}= t_{2} = t/2

The mean speed of car v_{m} = =15 km/h

A. 30 s

B. 20 s

C. 15 s

D. 10 s

Relative velocity of parrot w.r.t. train =10 + 20 = 30 m/s. Time taken by parrot to cross the train = 300 / 30 = 10 s

A. speed .

B. velocity.

C. acceleration.

D. time.

Slope of velocity-time graph represent the rate of change of velocity with time which is acceleration.

A. velocity is decreasing.

B. velocity is increasing.

C. speed is increasing.

D. velocity is zero.

A.

B.

C.

D.

velocity = dx/dt =constant (straight line)

acceleration, a = dv/dt = 0

i.e., position- time graph is a straight line.

A. vertically downward.

B. vertically upward.

C. at angle 45.

D. at angle 60.

A body`s acceleration due to gravity (g = 9.8 ms^{-2}) always acts in vertically downwards direction no matter whether it is rising or falling. When the body rises, the velocity decreases at the rate of 9.8 m/s every second. When the body is falling, its velocity increases at the rate of 9.8 m/s every second.

A. 0 m/s, 0 m/s.

B. 8 m/s, 0 m/s.

C. 0 m/s, 8 m/s.

D. 4 m/s, 4 m/s.

Average speed = total distance / total time = 200 / 25 = 8 m/s.

Net displacement = 0.

So, average velocity = 0.

A. - 12 ms^{-2}.

B. - 10 ms^{-2}.

C. - 8 ms^{-2}.

D. - 4 ms^{-2}.