Free Objective Test 02 Practice Test - 11th and 12th

Question 1

Two 20 kg cannon balls are chained together and fired horizontally with a velocity of 200 m/s from the top of a 30 m wall. The chain breaks during the flight of the cannon balls and one of them strikes the ground at t = 2 s, at a distance of 250 m from the foot of the wall, and 5 m to the right of the line of fire. Determine the position of the other cannon ball at that instant. Neglect the resistance of air.

A. (550, 20, -5)

B. (250, 20, 5)

C. (550, 10, -5)

D. (550, 20, 0)

SOLUTION

Solution : A

As no external force acts in z direction, hence z-coordinate of the centre of mas of the ball should be zero. To make z-coordinate zero other ball should fall symmetrically with respect to z-axis.
Hence, z-coordinate of other ball = –5 m.
The balls do not have any external force in x-direction. Hence in x-direction the centre of mass should move with constant velocity. X-coordinate of centre of mass at $t = 2 s = 200 \times 2 = 400 m$
Hence, $x_{C M} = \frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} + m_{2}}$
$400 = \frac{20 \times 250 + 20 x_{2}}{20 + 20}$
$x_{2} = 800 - 250 = 550 m$
Height fallen by centre of mass at t = 2s,
$h = \frac{1}{2} \times 10 \times (2)^{2} = 20 m$
Hence, y-coordinate of centre of mass $= 30 - 20 = 10 m$
Hence, $y_{C M} = \frac{m_{1} y_{1} + m_{2} y_{2}}{m_{1} + m_{2}}$
$10 = \frac{20 \times 0 + 20 \times y_{2}}{20 + 20}$
$\Rightarrow y_{2} = 20 m$

Question 2

Figure below shows the system is at rest initially with x = 0. A man and a woman both are initially at the extreme ends of the platform. The man and the woman start to move towards each other. Obtain an expression for the displacement 's' of the platform when the two meet in terms of the displacement ' $x_{1}$ ' of the man relative to the platform.

\frac{m_{1} x_{1} + m_{2} l}{m_{1} + m_{2} + m_{0}}

\frac{m_{1} x_{1} - m_{2} l}{m_{1} + m_{2} + m_{0}}

\frac{m_{1} x_{1} + m_{2} (l - x_{1})}{m_{1} + m_{2} + m_{0}}

\frac{m_{1} x_{1} - m_{2} (l - x_{1})}{m_{1} + m_{2} + m_{0}}

SOLUTION

Solution : D

As no external force is acting on the system, the displacement of the centre of mass should be zero. Let the displacement of the platform toward right be s,
$Δ {\to x}_{1} = Δ {\to x}_{1, p} + Δ {\to x}_{p} = (- x_{1} + s)$
$Δ {\to x}_{2} = Δ {\to x}_{2, p} + Δ {\to x}_{p} = (l - x_{1} + s)$
Hence,
$0 = \frac{m_{1} (- x_{1} + s) + m_{2} (l - x_{1} + s) + m_{0} s}{m_{1} + m_{2} + m_{0}}$
$s (m_{1} + m_{2} + m_{0}) - m_{1} x_{1} + m_{2} (l - x_{1}) = 0$
$s = \frac{m_{1} x_{1} - m_{2} (l - x_{1})}{m_{1} + m_{2} + m_{0}}$

Question 3

A block was released on the convex surface of hemispherical wedge as shown in the figure below. Determine the displacement of the wedge when the block reaches the angular position $θ$ .

\frac{m R c o s θ}{M + m}

to the right

\frac{m R s i n θ}{M + m}

to the right

\frac{m R c o s θ}{M + m}

to the left

\frac{m R s i n θ}{M + m}

to the left

SOLUTION

Solution : D

$m_{1} Δ {\to x}_{1} + m_{2} Δ {\to x}_{2} = 0$
$m (R s i n θ - x) - M x = 0$
$x = \frac{m R s i n θ}{m + M}$

Question 4

A block of mass 'm' is initially lying on a wedge of mass 'M' with an angle of inclination $θ$ , as shown in the figure. Calculate the displacement of the wedge when the block reaches bottom of the wedge.

\frac{m h}{M + m}

\frac{m h c o t θ}{M + m}

\frac{m h c o s θ}{M + m}

\frac{m h s i n θ}{M + m}

SOLUTION

Solution : B

$l = h c o s θ$
Let displacement of wedge be x towards right:
$Δ x_{C M} = \frac{m (l + x) + M x}{M + m}$
$0 = m l + (M + m) x$
$x = \frac{- m l}{M + m} = - \frac{m h c o t θ}{M + m}$

Question 5

A cart of mass 'M' is at rest on a frictionless horizontal surface and a pendulum bob of mass 'm' hangs from the roof of the cart. The string breaks, the bob falls on the floor, makes several collisions on the floor and finally lands up in a small slot made on the floor. The horizontal distance between the string and the slot is 'L'. Find the displacement of the cart during this process.

\frac{m l}{M + m}

to the right

\frac{m l}{M + m}

to the left

\frac{M l}{M + m}

to the right

\frac{M l}{M + m}

to the left

SOLUTION

Solution : B

Let displacement of the cart be $x_{c} = - x$

Displacement of the bob is

$x_{b} = (L - x)$
$(x_{C M})_{s y s t e m} = 0$
$M_{c a r t} x_{c a r t} + M_{b o b} x_{b o b} = 0$
$x = \frac{m L}{m + M}$

Question 6

Two blocks of masses $m_{1}$ and $m_{2}$ connected by a massless spring of spring constant 'k' rest on a smooth horizontal plane as shown in figure. Block 2 is shifted a small distance 'x' to the left and the released. Find the velocity of centre of mass of the system after block 1 breaks off the wall.

\frac{\sqrt{m_{2} K}}{m_{1} + m_{2}}

\frac{\sqrt{m_{1} K}}{m_{1} + m_{2}}

\frac{\sqrt{(m_{1} + m_{2}) K}}{m_{1} + m_{2}} x

D. zero

SOLUTION

Solution : A

If $m_{2}$ is shifted by a distance x and released, the mass $m_{1}$ will break off from the wall when the spring restores its natural length and $m_{2}$ will start going towards right. At the time of breaking $m_{1}$ , $m_{2}$ will be going towards right with a velocity v, which is given as
$\frac{1}{2} k x^{2} = \frac{1}{2} m_{2} v^{2} \Rightarrow v = (\sqrt{\frac{k}{m_{2}}}) x$
and the velocity of the centre of mass at this instant is
$v_{C M} = \frac{m_{1} \times 0 + m_{2} \times v}{m_{1} + m_{2}} = \frac{(\sqrt{m_{2} k}) x}{m_{1} + m_{2}}$

Question 7

Figure below shows a flat car of mass 'M' on a frictionless road. A small massless wedge is fitted on it as shown. A small ball of mass 'm' is released from the top of the wedge, it slides over it and falls in the hole at distance 'l' from the initial position of the ball. Find the distance the flat car moves till the ball gets into the hole.

\frac{m l}{M + m}

\frac{M l}{M + m}

\frac{m}{M} l

\frac{M}{m} l

SOLUTION

Solution : A

When the ball falls into the hole, with respect to the flat car, the ball travels as horizontal distance l. During this motion, to conserve momentum and to maintain the position of centre of mass, the car moves towards left, say by a distance x. Thus, the total distance travelled by the ball towards right is (l – x). As centre of mass remains at rest, the change in mass moments of the two (ball and car) about any point must be equal to zero. Hence
$m (l - x) = M x \Rightarrow m = \frac{m l}{M + m}$

Question 8

Two masses $m_{1}$ and $m_{2}$ placed at a distance 'l' apart, let the centre of mass of this system be at a point named C. If $m_{1}$ is displaced by $l_{1}$ towards C and $m_{2}$ is displaced by $l_{2}$ away from C, find the distance from C where the new centre of mass will be located.

\frac{m_{1} l_{1} + m_{2} l_{2}}{m_{1} + m_{2}}

\frac{m_{1} l_{1} - m_{2} l_{2}}{m_{1} + m_{2}}

l_{1} + l_{2}

l_{1} - l_{2}

SOLUTION

Solution : A

If $m_{1}$ and $m_{2}$ are placed at a distance l apart, their centre of mass will be located at a distance x from $m_{1}$ , where
$x = \frac{m_{2} l}{m_{1} + m_{2}}$
If $m_{1}$ is displaced by $l_{1}$ towards C and $m_{2}$ is displaced by $l_{2}$ away from C, the new centre of mass $C^{^{'}}$ now will be located at a distance $x^{'}$ from $m_{1}$ , where
$x^{^{'}} = \frac{m_{2} (l - l_{1} + l_{2})}{m_{1} + m_{2}}$
Displacement of the centre of mass is
$Δ x = x^{^{'}} + l_{1} - x = \frac{m_{1} l_{1} + m_{2} l_{2}}{m_{1} + m_{2}}$

Question 9

Figure below shows a fixed wedge on which two blocks of masses 2 kg and 3 kg are placed on its smooth inclined surfaces. When the two blocks are released from rest, find the acceleration of centre of mass of the two blocks.

\frac{\sqrt{31} g}{10}

B. g

C. g/2

D. g/5

SOLUTION

Solution : A

Block A slides with acceleration $\frac{g}{2}$ and block B slides with acceleration $\frac{\sqrt{3} g}{2}$ . Now the acceleration of centre of mass of the system of blocks A and B can be given both in x and in y directions as
$a_{x} = \frac{3 \times \frac{\sqrt{3} g}{2} c o s 60^{\circ} - 2 \times \frac{g}{2} s i n 30^{\circ}}{5} = \frac{\sqrt{3} g}{20}$
and $a_{y} = \frac{3 \times \frac{\sqrt{3} g}{2} s i n 60^{\circ} + 2 \times \frac{g}{2} s i n 30^{\circ}}{5} = \frac{11 g}{20}$
Thus acceleration of the centre of mass of the system is given as
$a_{C M} = \sqrt{a_{x}^{2} + a_{y}^{2}} = \sqrt{{(\frac{\sqrt{3} g}{20})}^{2} + {(\frac{11 g}{20})}^{2}} = \frac{\sqrt{31}}{10} g$

Question 10

Four particles, each of mass 'm', are placed at the corners of a square of side 'a' shown in the figure. The position vector of the centre of mass is

a (i + j)

\frac{a}{2} (i + j)

a (i - j)

\frac{a}{2} (i - j)

SOLUTION

Solution : B

The (x, y) co-ordinates of the masses at O, A, B and C respectively are
$(x_{1} = 0, y_{1} = 0), (x_{2} = a, y_{2} = 0), (x_{3} = a, y_{3} = a), (x_{4} = 0, y_{4} = a) .$
Therefore, the (x,y) co-ordinates of the centre of mass are
$x_{C M} = \frac{a}{2} a n d y_{C M} = \frac{a}{2}$ . The position vector of the centre of mass is
$\frac{a}{2} (i + j)$ , which is choice (b)

Question 11

A uniform rod of mass 'm' and length 'L' is tied to a vertical shaft. It rotates in horizontal plane about the vertical axis at angular velocity $ω$ . How much horizontal force does the shaft excert on the rod?

m w^{2} L

\frac{M w^{2} L}{2}

\frac{m w^{2} L}{4}

z e r o

SOLUTION

Solution : B

Let T be the force applied horizontally on the rod. The acceleration of centre of mass is $ω^{2} \frac{L}{2}$ , as it is moving in a circle of radius $\frac{L}{2}$ .
Using ${\to F}_{e x t} = M {\to a}_{c m} \Rightarrow T = M ω^{2} (\frac{L}{2})$
Thus, the horizontal force exerted is $\frac{1}{2} M ω^{2} L$

Question 12

Two blocks A and B each of equal masses 'm' are released from the top of a smooth fixed wedge as shown in the figure. The magnitude of acceleration of the centre of mass of the two blocks is ___.

SOLUTION

Solution :

Let us consider incline surfaces of wedge as x-axis and y-axis respectively. We can write accelerations of blocks A and B.
${\to a}_{A} = m g s i n 30^{\circ}^i a n d {\to a}_{B} = m g s i n 60^{\circ}^j$
The acceleration of centre of mass of two blocks can be written as
${\to a}_{a m} = \frac{m_{A} {\to a}_{A} + m_{B} {\to a}_{B}}{m_{A} + m_{B}} = \frac{m g s i n 30^{\circ}^i + m g s i n 60^{\circ}^j}{m + m}$
$= \frac{g s i n 30^{\circ}^i + m g s i n 60^{\circ}^j}{m + m}$
$= \frac{g}{4}^i + \frac{\sqrt{3} g}{4}^j \Rightarrow | {\to a}_{a m} | = \frac{g}{2}$

Question 13

Four Particles of masses 1 kg, 2 kg, 3 kg, and 4 kg are placed at the corners of a square of side 2 m in the x – y plane in anticlockwise sense. If the origin of the co-ordinate system is taken at the mass of 1 kg, the (x, y) co-ordinates of the centre of mass, expressed in 'metre' are

(1, \frac{7}{5})

(2, \frac{7}{5})

(3, \frac{7}{5})

(4, \frac{7}{5})

SOLUTION

Solution : A

The (x, y) co - ordinates of masses 1 kg, 2kg, 3kg and 4kg respectively are
$(x_{1} = 0, y_{1} = 0), (x_{2} = 2 m, y_{2} = 0), (x_{3} = 2 m, y_{3} = 2 m), (x_{4} = 0, y_{4} = 2 m) .$
The (x, y) co-ordinates of the centre of mass are
$x_{C M} = \frac{1 k g \times 0 + 2 k g \times 2 m + 3 k g \times 2 m + 4 k g \times 0}{1 k g + 2 k g + 3 k g + 4 k g} = 1 m$
$y_{C M} = \frac{1 k g \times 0 + 2 k g \times 0 + 3 k g \times 2 m + 4 k g \times 2 m}{1 k g + 2 k g + 3 k g + 4 k g} = \frac{7}{5} m$

Question 14

The motion of centre of mass of a system of two particles is unaffected by their internal forces

A. Irrespective of the actual directions of the internal forces

B. Only if they are along the line joining the particles

C. Only if they are at right angles to the line joining the particles

D. Only if they are obliquely inclined to the line joining the particles

SOLUTION

Solution : A

The correct choice is (a) because the centre of mass is affected only be external forces; not by internal forces.

Question 15

$lim x \to \infty \frac{(2 + x)^{40} (4 + x)^{5}}{(2 - x)^{45}}$

- 1

1

16

32

SOLUTION

Solution : A

$lim x \to \infty \frac{(2 + x)^{40} (4 + x)^{5}}{(2 - x)^{45}}$
$= lim x \to \infty \frac{x^{45} {(\frac{2}{x} + 1)}^{40} {(\frac{4}{x} + 1)}^{5}}{x^{45} {(\frac{2}{x} - 1)}^{45}}$
$= \frac{(1)^{40} (1)^{5}}{(- 1)^{45}}$
$= \frac{(1)^{45}}{(- 1)^{45}}$
$= - 1$

Free Objective Test 02 Practice Test - 11th and 12th

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

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