The sphere at A is given a downward velocity $$\displaystyle v_{0}$$ of magnitude $$5 m/s$$ and swings in a vertical plane at the end of a rope of length $$l=2 m$$ attached to a support at $$O$$. Determine the angle $$\displaystyle \theta $$ at which the rope will break, knowing that it can withstand a maximum tension equal to twice the weight of the sphere.

$$\displaystyle \sin ^{-1}\left ( \frac{1}{4} \right )$$

$$\displaystyle \sin ^{-1}\left ( \frac{1}{3} \right )$$

$$\displaystyle \sin ^{-1}\left ( \frac{1}{2} \right )$$

$$\displaystyle \sin ^{-1}\left ( \frac{3}{4} \right )$$

MCQ Questions for Class 11 Engineering Physics Laws Of Motion Quiz 12

The sphere at A is given a downward velocity

v_{0}

$\displaystyle v_{0}$ of magnitude

5 m / s

$5 m/s$ and swings in a vertical plane at the end of a rope of length

l = 2 m

$l=2 m$ attached to a support at

O

$O$ . Determine the angle

θ

$\displaystyle \theta$ at which the rope will break, knowing that it can withstand a maximum tension equal to twice the weight of the sphere.

0%
$\displaystyle \sin ^{-1}\left ( \frac{1}{4} \right )$
0%
$\displaystyle \sin ^{-1}\left ( \frac{1}{3} \right )$
0%
$\displaystyle \sin ^{-1}\left ( \frac{1}{2} \right )$
0%
$\displaystyle \sin ^{-1}\left ( \frac{3}{4} \right )$

Explanation

Given :

v_{o} = 5 m / s

$v_o =5 m/s$

l = 2

$l =2$ m

T = 2 m g

$T = 2mg$

From figure,

P M = l s i n θ

$PM = l sin\theta$

Circular motion equation at P :

\frac{m v^{2}}{l} = T - m g s i n θ

$\dfrac{mv^2}{l} = T - mgsin\theta$

m v^{2} = l (2 m g - m g s i n θ) = 2 (2 m \times 10 - m (10) s i n θ)

$mv^2 =l( 2mg - mgsin\theta) = 2 (2 m \times 10 - m(10) sin\theta)$

⟹

$\implies$

m v^{2} = m (40 - 20 s i n θ)

$mv^2 = m (40 - 20 sin\theta)$

Work-energy theorem :

W = Δ K . E

$W = \Delta K.E$

∴

$\therefore$

m g (P M) = \frac{1}{2} m v^{2} - \frac{1}{2} m v_{o}^{2}

$mg (PM) = \dfrac{1}{2}mv^2 - \dfrac{1}{2}mv_o^2$

2 m g (l s i n θ) = m (40 - 20 s i n θ) - m v_{o}^{2}

$2mg (l sin\theta) =m(40 - 20sin\theta) - mv_o^2$

2 m (10) (2 s i n θ) = m (40 - 20 s i n θ) - m (25)

$2 m (10) (2 sin\theta) = m (40 - 20 sin\theta) - m(25)$

⟹ s i n θ = \frac{1}{4}

$\implies sin\theta = \dfrac{1}{4}$

A weightless thread can withstand tension upto

30 N

$30\;N$ . A stone of mass

0.5 k g

$0.5\ kg$ is tied to it and is revolved in a circular path of radius

2 m

$2\ m$ in a vertical plane. If

g = 10 m / s^{2}

$g=10\ m/s^2$ , then the maximum angular velocity of the stone will be

0%
$5\ rad/s$
0%
$\sqrt{30}\ rad/s$
0%
$\sqrt{60}\ rad/s$
0%
$10\ rad/s$

Explanation

Given :

T_{m a x} = 30

$T_{max} =30$ N

m = 0.5 k g

$m = 0.5 kg$

r = 2

$r =2$ m

g = 10 m / s^{2}

$g = 10 m/s^2$

Tension in the string is maximum when the stone is at the lowest position of the vertical motion.

Let the maximum angular velocity be

w_{m a x}

$w_{max}$ .

Using circular motion equation :

m r w_{m a x}^{2} = T_{m a x} - m g

$mrw^2_{max} =T_{max} - mg$

∴

$\therefore$

0.5 \times 2 \times w_{m a x}^{2} = 30 - 0.5 \times 10

$0.5 \times 2 \times w^2_{max} =30 - 0.5 \times 10$

w_{m a x}^{2} = 30 - 5 = 25

$w^2_{max} = 30 - 5 =25$

⟹ w_{m a x} = 5

$\implies w_{max} = 5$

r a d / s

$rad/s$

A weightless thread can bear tension upto

3.7 k g

$3.7\;kg$ wt. A stone of mass

500 g m

$500\;gm$ is tied to it and revolved in a circular path of radius

4 m

$4m$ in a vertical plane. If

g = 10 m s^{- 2}

$g=10\;ms^{-2}$ , then the maximum angular velocity of the stone will be:

0%
$\;16\;rad/s$
0%
$\;\sqrt{21}\;rad/s$
0%
$\;2\;rad/s$
0%
$\;4\;rad/s$

Explanation

Given :

T_{m a x} = 3.7 k g w t = 37

$T_{max} =3.7 kg wt = 37$ N

m = 500

$m =500$ gm

= 0.5 k g

$= 0.5 kg$

r = 4

$r =4$ m

g = 10 m / s^{2}

$g = 10 m/s^2$

Tension in the string is maximum when the stone is at the lowest position of the vertical motion.

Let the maximum angular velocity be

w_{m a x}

$w_{max}$ .

Using circular motion equation :

m r w_{m a x}^{2} = T_{m a x} - m g

$mrw^2_{max} =T_{max} - mg$

∴

$\therefore$

0.5 \times 4 \times w_{m a x}^{2} = 37 - 0.5 \times 10

$0.5 \times 4 \times w^2_{max} =37 - 0.5 \times 10$

w_{m a x}^{2} = 16

$w^2_{max} =16$

⟹ w_{m a x} = 4

$\implies w_{max} = 4$

r a d / s

$rad/s$

Stone tied at one end of light string is whirled round a vertical circle. If the difference between the maximum and minimum tension experienced by the string wire is

2 k g w t

$2\ kg\ wt$ , then the mass of the stone must be

0%
$1\ kg$
0%
$6\ kg$
0%
$1/3\ kg$
0%
$2\ kg$

Explanation

Step 1: FBD [Ref. Fig.]

$\textbf{Step 1: FBD [Ref. Fig.]}$

Tension will be maximum at bottom and minimum at top

Step 2: Velocities at highest and lowest points

$\textbf{Step 2: Velocities at highest and lowest points}$

Minimum velocity required at bottom point to complete circle

= v

$= v$

v = \sqrt{5 g R}

$v = \sqrt{5gR}$

. . . . (1)

$....(1)$

Step 3: Calculations

$\textbf{Step 3: Calculations }$

For revolving in complete circle tension at highest point

T \geq 0

$T\geq 0$

\Rightarrow T_{min} = 0

$\Rightarrow T_{\min} = 0$

Applying Newton's second law on the stone along centripetal direction (Considering direction towards center as positive)

Σ F_{c} = m a_{c}

$\Sigma F_c = ma_c$

= m (\frac{v^{2}}{R})

$=m\left (\cfrac{v^2}{R} \right)$

Using equation

(1)

$(1)$ in the following

\Rightarrow T_{m a x} - m g = \frac{m v^{2}}{R} = 5 m g

$\Rightarrow T_{max} - mg = \dfrac{mv^2}{R} = 5 mg$

\Rightarrow T_{max} = 6 m g

$\Rightarrow T_{\max} = 6 mg$ [T is maximum at bottom]

So,

T_{max} - T_{min} = 6 m g

$T_{\max} - T_{\min} = 6mg$

\Rightarrow 6 m g = 2 g

$\Rightarrow 6mg = 2g$ Given:

T_{max} - T_{min} = 2 g

$T_{\max} - T_{\min} = 2g$

\Rightarrow m = \frac{2}{6} = \frac{1}{3} k g

$\Rightarrow m = \dfrac{2}{6} = \dfrac{1}{3} kg$

Hence

C

$C$ is the correct option.

2115133_280642_ans_40548d08555448c6834960c21299b13c.png

A uniform rod of mass m and length L is suspended with two massless strings as shown in the figure. If the rod is at rest in a horizontal position the ratio of tension in the two strings

T_{1}

$T_1$ /

T_{2}

$T_2$ is

0%
1 : 1
0%
1 : 2
0%
2 : 1
0%
4 : 3

Explanation

Vertical equilibrium

\sum F_{Y} = 0

$\sum F_Y=0$ ,

T_{1} + T_{2} = w

$T_1 + T_2 = w$ .........1

Taking Moments about point 'O',

Clockwise moments = Anticlockwise Moments;

T_{1} \times \frac{3 L}{4} = w \times \frac{L}{2}

$T_1 \times \dfrac{3L}{4}= w \times \dfrac{L}{2}$

∴ T_{1} = \frac{2 w}{3}

$\therefore T_1= \dfrac{2w}{3}$ .......2

Solving equations 1 and 2,

\Rightarrow T_{2} = \frac{w}{3}

$\Rightarrow T_2= \dfrac {w}{3}$ ,

∴ \frac{T_{1}}{T_{2}} = \frac{2}{1}

$\therefore \dfrac{T_1}{T_2}= \dfrac{2}{1}$

Two uniform rods of equal length but different masses are rigidly joined to form an L-shaped body, which is then pivoted about O as shown in the figure. If the system is in equilibrium in the shown configuration. Then the ratio M/m will be:

0%
$2$
0%
$3$
0%
$\sqrt{2}$
0%
$\sqrt{3}$

Explanation

Taking Moments about point 'O'

clockwise moments = Anticlockwise moments,

M g \times \frac{l}{2} s i n (30^{o})

$Mg \times \dfrac{l}{2} sin(30^o)$ =

m g \times \frac{l}{2} s i n (60^{o})

$mg \times \dfrac{l}{2} sin(60^o)$ ,

\Rightarrow \frac{M}{2} = m \frac{\sqrt{3}}{2}

$\Rightarrow \dfrac{M}{2}= m \dfrac{\sqrt3}{2}$

∴ \frac{M}{m} = \sqrt{3}

$\therefore \dfrac{M}{m}= \sqrt{3}$

The vertical section of a road over a canal bridge in the direction of its length is in the form of circle of radius

8.9 m

$8.9\;m$ . Then the greatest speed at which the car can cross this bridge without losing contact with the road at its highest point, the centre of gravity of the car being at a height

$h=1.1\;m$ from the ground (Take

$g=10m/sec^2$ )

0%
$\;5\;m/sec$
0%
$\;10\;m/sec$
0%
$\;15\;m/sec$
0%
$\;20\;m/sec$

Explanation

Given : Radius of the circular arc

$r =8.9$ m

$g = 10 m/s^2$

Heigh of centre of gravity of car above the road

$h =1.1$ m

Centre of gravity of the car is revolving in a circular motion.

$\therefore$ Distance of centre of gravity of car and the centre of circle

$l = r+h = 8.9 + 1.1 =10$ m

Using

$\dfrac{mv^2}{l} = mg - N$

For maximum speed of car

$(v_{max})$ ,

$\implies N = 0$

$\therefore$

$\dfrac{mv_{max}^2}{l} = mg$

$\dfrac{m v^2 _{max}}{10} = m \times 10$

$\implies v_{max} =10$

$m/s$

A body crosses the topmost point of a vertical circle with critical speed. What will be its centripetal acceleration when the string is horizontal. :-

0%
$\;g$
0%
$\;2g$
0%
$\;3g$
0%
$\;6g$

Explanation

The critical speed of the body at the topmost point is equal to

$\sqrt{gR}$ i.e

$v = \sqrt{gR}$

Let the velocity of the body when the string is horizontal be

$v_2$

Using work-energy theorem :

$W = \Delta K.E$

$\therefore$

$mgR = \dfrac{1}{2} m v_2^2 - \dfrac{1}{2}mv^2$

$mgR = \dfrac{1}{2} m v_2^2 - \dfrac{1}{2}m(gR)$

$\implies v_2^2 = 3gR$

Centripetal acceleration

$a_r = \dfrac{v_2^2}{R} = 3g$

A metal bar

$70\ cm$ long and

$4.00\ kg$ in mass supported on two knife-edges placed

$10\ cm$ from each end. A

$6.00\ kg$ load is suspended at

$30\ cm$ from one end. Find the reactions at the knife-edges. (Assume the bar to be of uniform cross section and homogeneous.)

0%
$45\ N$ and $43\ N$
0%
$50\ N$ and $35\ N$
0%
$55\ N$ and $43\ N$
0%
$54\ N$ and $30\ N$

Explanation

Figure shows the rod AB, the positions of the knife edges

$K_1$ and

$K_2$ , the centre of gravity of the rod at G and the suspended load at P.
Note the weight of the rod W acts as its centre of gravity G. The rod is uniform in cross section and homogenous

$;$ hence G is at the centre of the rod

$; AB=70$ cm.

$AG=35$ cm,

$AP=30$ cm,

$PG=5$ cm,

$AK_1=BK_2=10$ cm and

$K_1G=K_2G=25$ cm. Also,

$W=$ weight of the rod

$=4.00$ kg and

$W_1=$ suspended load

$=6.00$ kg

$;$

$R_1$ and

$R_2$ are the normal reactions of the support at the knife edges.
For translational equilibrium of the rod,

$R_1+R_2-W_1-W=0$ ....

$(i)$
Note

$W_1$ and

$W$ act vertically down and

$R_1$ and

$R_2$ act vertically up.
For considering rotational equilibrium, we take moments of the forces. A convenient point to take moments of the forces is G. The moments of

$R_2$ and

$W_1$ are anticlockwise

$(+ve)$ , whereas the moment of

$R_1$ is clockwise

$(-ve)$ .
For rotational equilibrium,

$-R_1(K_1G)+W_1(PG)+R_2(K_2G)=0$

$(ii)$
It is given that

$W=4.00$ g N and

$W_1=6.00$ g N, where

$g=$ acceleration due to gravity. We take

$g=9.8m/s^2$ .
With numerical values inserted, from

$(i)$

$R_1+R_2-4.00g-6.00g=0$
or

$R_1+R_2=10.00g N$

$(iii)$

$=98.00N$
From

$(ii), -0.25 R_1+0.05W_1+0.25R_2=0$
or

$R_1-R_2=1.2g$

$N=11.76N$

$(iv)$
From

$(iii)$ and

$(iv)$ ,

$R_1=54.88N$ ,

$R_2=43.12N$
Thus the reactions of the support are about

$55N$ at

$K_1$ and

$43N$ at

$K_2$ .

A weightless thread can bear tension upto

$37 N$ . A stone of mass

$500 g$ is tied to it and revolved in a circular path of radius

$4 m$ in a vertical plane. If

$g=10 {ms}^{-2}$ , then the maximum angular velocity of the stone will be:

0%
$2\ rad\ {s}^{-1}$
0%
$4\ rad\ {s}^{-1}$
0%
$8\ rad\ {s}^{-1}$
0%
$16\ rad\ {s}^{-1}$

Explanation

Maximum tension in the string will be at the lowest point of the circular path, where the tension will support the weight as well as provide the centripetal force for circular motion.

So, we have:

$T=mg+ml{ \omega }^{ 2 }$

$\Rightarrow ml{ \omega }^{ 2 }=T-5$

For maximum angular speed, the tension must be

$37N$

So, the maximum angular speed is

$\sqrt { \dfrac { T-5 }{ ml } } =\sqrt { \dfrac { 37-5 }{ 0.5\times 4 } } = 4 rad/s$

Thus,

${ \omega }_{ max }= 4 rad/s$

Two blocks of mass 5 kg and 10 kg are kept on a rough horizontal surface as shown in figure A force F is applied on upper block of mass 10 kg Now choose the correct statement(s) regarding the system

0%
If magnitude of applied force is 2N then frictional force between the blocks is also 2N
0%
The acceleration of 10 kg block is 2 $\displaystyle m/s^{2}$ when applied force is 30 N
0%
The minimum value of F for which 5 kg block begin to slide if coefficient of friction between blocks is changed to 0.5 is 45 N
0%
5 kg block will never move on the ground for any value of F if coefficient of friction between blocks is 0.1 and between block and surface is 0.3

Explanation

For 10kg block, f is frictional force

$f_{1max} = \mu_1 \mu_2 = \mu_1 (10g) = 0.1(10 \times 10) = 10N$

i) if $F = 2N$ then acting friction

$f_1 = F = 2N$

(A) is correct.

for 5 kg block;

$f_{2max} = \mu (10 + 5)g = 0.3 \times 15 \times 10= 45 N$

ii) if $F = 30N$ then;

$F - F_{1max}= 10a$

$10a = 30 - 10 =20$

$a =2 m/s^2$

(B) is correct.

iii) if $\mu_1$ changed to 0.5;

then $f_{1max} = \mu_1 (10g) = 0.5 \times 10 \times 10 = 50N$

In the case acting friction will be $F_{2max}$

Since,

$F_{1max} = F_{2max}$

Block 10kg and 5kg both move together then $F_{min}$ to begin is

$F_{min} - F_{max} = 0$

$F_{min} = F_{2max} = 45 N$

iv) if $\mu = 0.1$ , then $f_{max} = 10 N$ and,

$f_{2max} = 45 N$

$f_{2max} > f_{1max}$ always for any any value of $N$ . Hence, 5kg block never move on the ground.

So, option (A), (B), (C), and (D) are correct.

If a particle of mass

$M$ is tied to a light inextensible string fixed at point

$P$ and particle is projected at A with velocity

$V_A\, =\, \sqrt{4 gL}$ as shown. Find tension in the string at

$B$ . (Assume particle is projected in the vertical plane.)

0%
$Mg$
0%
$3Mg$
0%
$2Mg$
0%
$7Mg$

Explanation

Taking point A as reference where gravitational potential energy is zero,

Total energy of particle at point A is,

$E_A = \dfrac{1}{2}MV_A^2 = 2MgL$

Let velocity at point B as

$V_B$ .

So, Total energy at point B is,

$E_B = mgL + \dfrac{1}{2}MV_B^2$

By conservation of energy,

$E_A=E_B$

$\implies \dfrac{1}{2}MV_B^2 = mgL\implies \dfrac{MV_B^2}{L}=2Mg$

So centripetal force point B is

$2Mg$

At point B,

$T+Mg=\dfrac{MV_B^2}{L}=2Mg\implies T=Mg$

So Tension in the string is

$Mg$ , at point B.

Assertion: A quick collision between two bodies is more violent than a slow collision; even when the initial and final velocities are identical.
Reason: The momentum is greater in first case

0%
If both assertion and reason are true but the reason is the correct explanation of assertion.
0%
If both assertion and reason are true but the reason is not the correct explanation of assertion.
0%
If assertion is true but reason is false.
0%
If both the assertion and reason are false.
0%
If reason is true but assertion is false.

A car taken a turn along a curved road of radius

$40$ m and the angle of banking of the road is

$\displaystyle {45} ^ {o}$ . At what maximum speed can the car be driven so that it may not skid.

$\displaystyle (g\ = {10\ ms} ^{-2}$ ).

0%
$\displaystyle {20\ ms} ^{-1}$
0%
$\displaystyle {10\ ms} ^{-1}$
0%
$\displaystyle {5\ ms} ^{-1}$
0%
$\displaystyle {1\ ms} ^{-1}$

Explanation

For banking of road:

$V^2\le rg\tan \theta$

$\implies V \le \sqrt{rg\tan \theta}$

$\implies V\le \sqrt{40\times 10\times \tan 45^o}$

$\implies V\le 20$

$\implies V_{max}=20$

${m/s}$

A small sphere is given vertical velocity of magnitude

$v_{0} = 5m/s$ and it swings in a vertical plane about the end of massless string. The angle

$\theta$ with the vertical at which string will break, knowing that it can withstand a maximum tension equal to twice the weight of the sphere, is

$[g = 10 m/s^{2}]$ .

0%
$\cos^{-1}\left (\dfrac {2}{3}\right )$
0%
$\cos^{-1}\left (\dfrac {1}{4}\right )$
0%
$60^{\circ}$
0%
$30^{\circ}$

A uniform meter sticks of mass 200 g is suspended from the ceiling through two verticle strings of equal lengths fixed at the ends. A small object of 20 g is placed on the stick at a distance of 70 cm from the left end. Find the tension in the two strings.

0%
$T_1=1.06N, T_2=1.14N$
0%
$T_1=1.03N, T_2=1.17N$
0%
$T_1=1.10N, T_2=1.10N$
0%
$T_1=1.00N, T_2=1.20N$

Explanation

C is the centre of mass of the stick.

For translation equilibrium

$T_1 + T_2 = 220g--------(1)$

Here,g is acceleration due to gravity

$=1000cm/s^2$

For rotation equilibrium torque about centre of mass shall be zero or sum of clockwise torque is equal to sum anti clockwise torque.

So, AC=CB=50cm

$T_1 \times 50 + 20g \times (70-50)=T_2\times 50$

$\implies T_2 -T_1 = 8g---------(2)$

So,from (1) & (2) we get

$T_2=114g=1.14\times 10^5 dyne =1.14N$

$T_1=106g=1.06\times 10^5 dyne = 1.06N$

If the beads are released from rest, such that the bead

$A$ is at the highest position of ring at time

$t = 0$ , the magnitude of acceleration of the bead

$B$ in the vertical direction at the same instant is

0%
Zero
0%
$\dfrac {g}{4}$
0%
$\dfrac {g}{2}$
0%
$g$

Explanation

For A that is rest at top

$T\cos { 45° } =mg\\ T=\sqrt { 2 } mg$

at B.

$T\sin { 45° } -mg=m0\\ \sqrt { 2 } mg\times \dfrac { 1 }{ \sqrt { 2 } } -mg=mg\\ 0=mg\\ g=O$

A mass tied to a string moves in a vertical circle and at the point

$P$ its speed is

$5m/s$ . At the point

$P$ the string breaks. The mass will reaches height above

$P$ of nearly

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

0%
$1m$
0%
$0.5m$
0%
$1.27m$
0%
$1.25m$

A uniform rod of length

$L$ rests against a smooth wall as shown in figure. Find the friction coefficient between the ground and the lower end if the minimum angle that the rod can make with the horizontal is

$\theta$ .

0%
$\dfrac{L\cos^2\theta \sin\theta}{2h-l\cos\theta \sin^2\theta}$
0%
$\dfrac{L\cos\theta \sin^2\theta}{2h-l\cos\theta \sin^2\theta}$
0%
$\dfrac{L\cos^2\theta \sin\theta}{2h-l\cos^2\theta \sin\theta}$
0%
$\dfrac{L\cos\theta \sin^2\theta}{2h-l\cos^2\theta \sin\theta}$

Explanation

Figure(b) gives the forces .

$R_2=$ Reactional force on rod by ground

$R_1=$ Reactional force on rod by wall

$f=$ frictional force

$=\mu R_2$

$\theta$ is the maximum angle the rod can make with horizontal.

Equating the forces on verticle direction we get

$R_2=mg-R_1cos\theta--------(1)$

Equating the forces on horizontal direction we get

$R_1 sin\theta = f =\mu R_2$

Now the torque about point B should be balanced

$R_1 cos\theta \times AB + R_1 Sin\theta \times h = mg\cfrac{l}{2}cos\theta$

$\implies R_1 h[\cfrac{cos^2\theta +sin^2\theta}{sin\theta}]=\cfrac{mgl}{2}cos\theta$

$\implies R_1=\cfrac{mgl}{2h} cos\theta sin\theta$

$\implies R_1 cos\theta = \cfrac{mgl}{2h}cos^2\theta sin\theta$

So,

$R_2 = mg-\cfrac{mgl}{2h}cos^2\theta sin\theta = \cfrac{mg}{2h} (2h-lcos^2\theta sin\theta)$

And

$\mu = \cfrac{R_1 sin\theta}{R_2} = \cfrac{\cfrac{mgh}{2h} cos\theta sin^2\theta}{\cfrac{mg}{2h} (2h-lcos^2\theta sin^2 \theta)}$

$\mu =\cfrac{lcos\theta sin^2 \theta}{2h-lcos^2\theta sin\theta}$

An object is said to be in equilibrium when:
there is no net force acting on the object
the total clockwise moments about any point is equal to the total anti-clockwise moments about the same point
the object moves with constant velocity

0%
$1$ only
0%
$2$ only
0%
$1$ and $2$ only
0%
$1, 2$ and $3$

Explanation

A body is said to be in equilibrium if it in translational equilibrium as well as rotational equilibrium.

For translational equilibrium, net force acting on the body must be zero i.e.

$F_{net} = 0$

For rotational equilibrium, net torque acting on the body must also be zero or in other words, total clockwise moments about any point is equal to the total anti-clockwise moments about the same point.

A particle of mass 'm' strikes a smooth stationary wedge of mass M with a velocity

$v_0$ , at an angle

$\theta$ with horizontal if the collision is perfectly inelastic, the impulse on the wedge is:

0%
$\dfrac{Mmv_0}{M+m}$
0%
$Mv_0$
0%
$\dfrac{Mmv_0cos\theta}{M+m}$
0%
$\dfrac{Mmv_0sin\theta}{M+m}$

Explanation

Initial momentum along x axis

$= mv_0 \cos\theta$

Final momentum

$= (M + M)v$

$v = \cfrac{mv_0\cos\theta}{M + m}$

Impulse = Change in momentum

$= \cfrac{Mmv_0\cos\theta}{M + m}$

Two blocks are connected by a massless string that passes over a frictionless peg as shown in fig. One end of the string is attached to a mass

$m_1 = 3kg$ , i.e. a distance

$R = 1.20 m$ from the peg. The other end of the string is connected to a block of mass

$m_2 = 6 kg$ resting on a table. When the 3 kg block be released at

$\displaystyle \theta = \frac{\pi}{k}$ , the 6 kg block just lift off the table? Find the value of k.

0%
$2$
0%
$3$
0%
$4$
0%
$6$

Explanation

Tension in the string will be maximum when

$m_1$ is at lowest position. For

$m_2$ to just lift, tension in the string at this position should equal the weight of

$m_2$ at this position.

Using conservation of energy,

$\dfrac{m_1 R^2 \omega^2}{2} = m_1 g R (1-\cos \theta)$

$m_1 \omega^2 R = 2m_1 g (1 - \cos \theta) \quad ........(i)$

From the FBD of

$m_1$ ,

$T_1 - m_1g = m_1 \omega^2 R \quad ...........(ii)$

From

$(i)\ \& \ (ii),$

$T_1 = m_1g + 2m_1g (1-\cos \theta)$

$T_1 = m_1g (3-2\cos\theta) \quad .........(iii)$

From the FBD of

$m_2$ ,

$T_2 + N_2 = m_2 g \quad ........(iv)$

When the block just leaves the surface,

$N_2 = 0 \quad .....(v)$

Also, tension throughout the string is constant,

$T_2=T_1 \quad ........(vi)$

Using

$(iii), \ (iv), \ (v) \ \& \ (vi),$

$m_2g = m_1g(3-2cos\theta)$

$\cos \theta = \dfrac{3 - m_2/m_1}{2}$

Substituting values,

$\cos \theta = 1/2$

$\theta = \pi/6$

$\implies k = 6$

A massless meter stick is in equilibrium about its center point.
If a mass,

${m}_{1}=50 g$ , is placed at the

$30 cm$ mark and a mass,

${m}_{2}=75 g$ , is placed at the

$70 cm$ mark, where must a mass,

${m}_{3}=50 g$ , be placed to keep the system in equilibrium?

0%
$10 cm$
0%
$40 cm$
0%
$50 cm$
0%
$60 cm$
0%
$90 cm$

Explanation

Given :

$m_1 = 50$ gram

$m_2= 75$ gram

$m_3 = 50$ gram

As the meter stick is in rotational equilibrium, thus net torque about point C must be equal to zero i.e

$\tau_c = 0$

$\therefore$

$m_1 g (20) + m_3 g (x) - m_2 g (20) =0$

$50 \times 20 + 50 x - 75 \times 20 = 0$

$\implies x =10$ cm

Hence the mass

$m_3$ must be palced at position

$X_3 = 50 - 10 = 40$ cm

Two masses

$A$ and

$B$ of

$10Kg$ and

$5Kg$ respectively are connected with a string passing over a frictionless pulley fixed at the corner of a table as in figure. The coefficient of friction of

$A$ with the table is

$0.2$ . The minimum mass of

$C$ that may be placed on

$A$ to prevent it from moving is equal to:

0%
$20Kg$
0%
$15Kg$
0%
$10Kg$
0%
$5Kg$
0%
$0Kg$

$"V"$ shaped rigid body has two identical uniform arms. What must be the angle between the two arms so that when the body is hung from one end, the other arm is horizontal?

0%
$\cos^{-1}(1/3)$
0%
$\cos^{-1}(1/2)$
0%
$\cos^{-1}(1/4)$
0%
$\cos^{-1}(1/6)$

A particle of mass

$m$ strikes a wall elastically with speed

$v$ at an angle

${30}^{o}$ with the wall as shown in the figure. The magnitude of impulse imparted to the ball by the wall is:

0%
$mv$
0%
$\cfrac{mv}{2}$
0%
$2mv$
0%
$\sqrt {3} mv$

Explanation

Before the particle strike,

Linear momentum perpendicular to the wall

$= mvsin(30^o)= \dfrac{mv}{2}$

Linear momentum along to the wall

$= mvcos(30^o)$

After the particle strike,

Linear momentum perpendicular to the wall

$= -mvsin(30^o)= -\dfrac{mv}{2}$

Linear momentum along to the wall

$= mvcos(30^o)$

Impulse=change in linear momentum,

Linear Momentum changed only perpendicular to the wall.

$\Rightarrow Impulse= -\dfrac{mv}{2}-\dfrac{mv}{2}=-mv$

Magnitude of Impulse is

$mv$

A stone of mass

$1\ kg$ tied to a light inextensible string of length

$L = \dfrac{10}{3}\ m$ , whirling in a circular path in a vertical plane. The ratio of maximum tension in the string to the minimum tension in the string is

$4$ . If g is taken to be

$10\ m/s^2$ the speed of the stone at the highest point of the circle is

0%
$10\ m/s$
0%
$5\sqrt{2}\ m/s$
0%
$10\sqrt{3}\ m/s$
0%
$20\ m/s$

Explanation

$The\quad ratio\quad to\quad the\quad mamimum\quad { T }_{ 2 }{ \quad to\quad the\quad minimum\quad tension\quad T }_{ 1 }\quad is\quad -\\ \frac { { T }_{ 2 } }{ { T }_{ 1 } } =4\\ { T }_{ 2 }=4{ T }_{ 1 }\\ Now\quad the\quad difference\quad between\quad the\quad two\quad 6mg;\\ Hence\quad we\quad have\quad \\ { T }_{ 2 }-{ T }_{ 1 }=6mg\\ \therefore 4{ T }_{ 1 }-{ T }_{ 1 }=6mg\\ 3{ T }_{ 1 }=6mg\\ { T }_{ 1 }=2mg\\ Now\quad tension\quad at\quad the\quad top\quad of\quad the\quad circle\quad is-\\ { T }_{ 1+mg }=\frac { { mv }_{ 1 }^{ 2 } }{ r } =\frac { { mv }_{ 1 }^{ 2 } }{ L } \\ 2mg+mg=\frac { { mv }_{ 1 }^{ 2 } }{ \frac { 10 }{ 3 } } \\ { v }_{ 1 }^{ 2 }=3g\frac { 10 }{ 3 } =10g\\ { v }_{ 1 }=\sqrt { 10g } \\ =\sqrt { 10\times 10 } =10m/s$

A uniform rod AB mass m= 1.12 kg and length l= 100 cm is placed on a sharp support O such that AO = a = 40 cm and OB = b = 60 cm. To keep the rod horizontal, its end A is tied with a thread. Calculate reaction of support O on the rod when the thread is burnt. ( g 10

$m{ s }^{ -1 }$ )

0%
40 N
0%
10 N
0%
15.5 N
0%
20 N

A uniform rod of mass

$M$ and length

$L$ is pivoted at one end such that it can rotate in a vertical plane. There is negligible friction at the pivot. The free end of the rod is held vertically above the pivot and then released. The angular acceleration of the rod when it makes an angle

$\theta$ with the vertical is

0%
$g\sin { \theta }$
0%
$\cfrac { g }{ L } \sin { \theta }$
0%
$\cfrac { 3g }{ 2L } \sin { \theta }$
0%
$6gL\sin { \theta }$

Explanation

$\tau_o= I\alpha$

Moment of inertia of rod about O,

$I= \dfrac{ML^2}{3}$

$\tau_o= Mg\dfrac{L}{2}sin\theta$

thus

$Mg\dfrac{L}{2}sin\theta= \dfrac{ML^2}{3} \alpha$

$\implies \alpha= \dfrac{3g}{2L}sin\theta$

1537040_984327_ans_bf6e8797d20446bf8f31cf7fb13f3e83.PNG

Maximum force exerted by the bat on the ball is:

0%
$2500 N$
0%
$5000 N$
0%
$7500 N$
0%
$1250 N$

Explanation

Mass of balls

$\Rightarrow { m }_{ b }=250g$

${ u }_{ { b }_{ 1 } }=30m/s$

abler strike

${ u }_{ { b }_{ 2} }=50m/s$

$\Rightarrow \triangle u=20m/s$

$\Rightarrow \dfrac{ \triangle u}{t}=20\times 10^3m/s$

$\Rightarrow F=m\times \dfrac{ \triangle u}{t}=20\times 10^3\times .250$

$=5000N$

Hence, the answer is

$5000N.$

A uniform hollow hemisphere of mass m and radius r is released from rest on a smooth horizontal surface with its open face vertical initially. Find out the Maximum normal reaction between hemisphere and ground during motion is

0%
$\dfrac { 10mg }{ 5 }$
0%
$\dfrac { 9mg }{ 5 }$
0%
$\dfrac { 11mg }{ 5 }$
0%
$\dfrac { 12mg }{ 5 }$

$1\ kg$ stone at the end of

$1\ m$ long string is whirled in a vertical circle at a constant speed of

$4\ ms^{-1}$ . The tension in the string is 6 N when the stone is

0%
At the top of the circle
0%
At the bottom of the circle
0%
Half way down
0%
None of above

Explanation

$\textbf{Step 1 - Draw FBD }$

$\textbf{Step 2 - Apply Newton second law -}$

$\sum F=ma$

along the radially -

$F_{net} = Ma_{R}$

$T - mg\cos \theta = \dfrac {mv^{2}}{R}$

$T = mg \cos \theta + \dfrac {mv^{2}}{R}$

$\textbf{Step3:Put values to obtain answer}$

As given that -

$T = 6N, m = 1\ kg, g = 10\ m/s^{2}, v = 4\ m/s$ and

$R = 1\ m$

So, put these all values in the above equation -

$6 = 1\times 10\times \cos \theta + \dfrac {1\times (4)^{2}}{1}$

$6 = 10\cos \theta + 16$

$10\cos \theta = -10$

$\cos \theta = -1$

$\theta = 180^{\circ}$

Means, stone is to be on top of the circle.

So, the tension in the string is

$6N$ when -

the stone is at the top of circle.

Hence, option (A) is correct.

Figure shows the variation of force acting on a body with time. Assuming the body to start from rest, the variation of its momentum with time is best represented by which plot?

A uniform rod of length

$4L$ and mass

$M$ is suspended form a horizontal roof by two light strings of length

$L$ and

$2L$ as shown. Then the tension in the left string of length

$L$ is:

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

Explanation

$AC = 4L$

$BC = L$

$AB = \sqrt {A{C^2} - B{C^2}}$

$= \sqrt {{{\left( {4L} \right)}^2} - {L^2}}$

$= \sqrt {16{L^2} - {L^2}}$

$= \sqrt {15{L^2}}$

$AB = \sqrt {15} L$
Also,

$\cos \theta = \dfrac{{AB}}{{AC}} = \dfrac{{AE}}{{AD}}$

$= \dfrac{{\sqrt {15} L}}{{4L}} = \dfrac{{AE}}{{2L}}$

$AE = \dfrac{{\sqrt {15} }}{4} \times 2L$

$AE = \dfrac{{\sqrt {15} }}{2}L$
taking moment at

$A$

${T_1} \times AB = Mg \times AE$

${T_1} \times \sqrt {15} L = Mg \times \dfrac{{\sqrt {15} }}{2}L$

${T_1} = \dfrac{{Mg}}{2}$

If a street light of mass

$M$ is suspended from the end of uniform rod of length

$L$ in the different possible patterns as shown in figure, then

0%
Pattern $A$ is more sturdy
0%
Pattern $B$ is more sturdy
0%
Pattern $C$ is more sturdy
0%
none of these

Explanation

$\text { Torque croated due to weight of streat light remains } \\$

$\text { same in all the three cases. } 9 t \text { is balanced by } \\$

$\text { torque created by tension in the string. So if } \gamma \\$

$\text { be the torque created by weight of lamp and } T \text { ' } \\$

$\text { be tension in the string and ' } d \text { ' be perpendicular } \\$

$\text { distance of cable from the axis thou, }$

$\tau=T \cdot d$

$\text { Tension will be least for laxgest'd'. This is } \\$

$\text { in pattern } A \text { is more sturdy. } \\$

For stable equilibrium, which of the following is/are true?

0%
$\cfrac { dU }{ dx } =0$ and $\cfrac { { d }^{ 2 }U }{ d{ x }^{ 2 } } >0$
0%
$\cfrac { dU }{ dx } =0\quad$ only
0%
$\cfrac { dU }{ dx } =0$ and $\cfrac { { d }^{ 2 }U }{ d{ x }^{ 2 } } < 0$
0%
None of these

What angle a bicycle and its rider must make with the vertical when travelling at

$54\ km/hr$ around a horizontal curve of radius

$40\ m$ ?

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

0%
$\sin^{-1}\ (0.56)$
0%
$\cot^{-1}\ (1.78)$
0%
$\cos^{-1}\ (0.56)$
0%
$\tan^{-1}\ (0.56)$

A string wraps around a fat pipe as a bob attached to the string is made to move in a circular path in the horizontal. Assuming the speed is somehow held constant as the radius diminishes due to the wrapping, how will the centripetal force change?

0%
It will stay the same.
0%
It will diminish.
0%
It will increase
0%
None of the above (this is a jokeits got to be one of the three above)

A small block of mass m is pushed on a smooth track from position A with a velocity

$2\sqrt5$ times the minimum velocity required to reach point D. The block will leave the contact with track at the point where normal force between them becomes zero.
At what angle

$\theta$ with horizontal does the block gets separated from the track?

0%
$sin^{-1}(\frac{1}{3})$
0%
$sin^{-1}(\frac{3}{4})$
0%
$sin^{-1}(\frac{2}{3})$
0%
never leaves contact with the track

A small block of mass m is pushed on a smooth track from position A with a velocity

$2\sqrt5$ times the minimum velocity required to reach point D. The block will leave the contact with track at the point where normal force between them becomes zero.
When the block reaches point B, what is the direction (in terms of angle with horizontal) of acceleration of the block?

0%
$tan^{-1}(\frac{1}{2})$
0%
$tan^{-1}(2)$
0%
$sin^{-1}(\frac{2}{3})$
0%
The block never reaches point B.

A rod of mass

$m$ and length

$l$ in placed on a smooth table. An another particle of same mass

$m$ strikes the rod with velocity

${v}_{0}$ in a distance perpendicular to the rod at distance

$x(</2)$ from its centre. Particle sticks to the rod. Let

$\omega$ be the angular speed of system after collision, then find maximum possible value of impulses (by varying

$x$ ) that can be imparted to the particle during collision. Particle still sticks to the rod.

0%
$\dfrac { m{ v }_{ 0 } }{ 2 }$
0%
$\dfrac { 2m{ v }_{ 0 } }{ 3 }$
0%
$\dfrac { 3m{ v }_{ 0 } }{ 4 }$
0%
$\dfrac { 4mv }{5}$

A cyclist going around a circular road of radius

$10m$ is observed to be bending inward 30

$^\circ$ with vertical. Frictional force acting on the cyclist is (Given:

$g = 10 m/s^2$ , mass of the cyclist is

$90$ Kg)

0%
$532 N$
0%
$800N$
0%
$1559 N$
0%
$520 N$

A solid sphere is placed on a horizontal plane. A horizontal impulse

$I$ is applied at a distance

$h$ above the central line as shown in the figure. Soon after giving impulse the sphere starts rolling

0%
$\cfrac{1}{2}$
0%
$\cfrac{2}{5}$
0%
$\cfrac{1}{4}$
0%
$\cfrac{1}{5}$

A mass

$m$ is revolving in a vertical circle at the end of a string of length

$20\ cm$ . By how much times does the tension of the string at the lowest point exceed the tension at the topmost point-

0%
$2\ mg$
0%
$4\ mg$
0%
$6\ mg$
0%
$8\ mg$

Explanation

Let tension

$T_1$ at top most point is given by

$T_1 + mg = \dfrac{mV^2_1}{l}$ ...(1)

similarly, the tension

$T_2$ at the lowest point is given by

$T_2 -mg = \dfrac{mV_2^2}{l}$ ...(1)

here, tension acting in words while balancing centrifugal force and weight of the body acting outwards

$T_2 - T_1 = \dfrac{mv^2_2}{l} +mg - \dfrac{mv_1^2}{l} +mg$

$\dfrac{m}{l} (v^2_2 -V^2_1) + 2mg$

we know from kilometres

$v^2_1 = v^2_2 - 2g(2l)$

$T_2 T_1 = \dfrac{m}{l} [2g(2l)] +2mg$

$= 6mg$

Hence (C) is correct answer

A ball of mass 0.2 Kg is dropped from a certain height above the ground.It bounces back after an elastic collision with the floor.If the speed with which the ball strikes the ground is 10m/s, then the impluse imparted by the ball on the floor is:

0%
12 Kg m/s
0%
8 Kg m/s
0%
4 Kg m/s
0%
2 Kg m/s

The motion of particle of mass m is given by x = 0 for t < 0 s , x (t) = A sin4p t for 0 < t < (1/4) s ( A > 0), and x = 0 for t > (1/4) s. Which of the following statement is true?

0%
The force at t = (1/8) s on the particle is $-a6\pi^2 \ A \ m$
0%
The particle is acted upon by on impulse of magnitude $4\pi^2 \ A \ m$ at t = 0 s and t = ( 1/4) s.
0%
The particle is not acted upon by any force.
0%
The particle is not acted upon by a constant force.
0%
There is no impulse acting on the particle.

As shown in the given figure the ball is given sufficient velocity at the lowest point to complete the circle. Length of string is

$1m$ . Find the tension in the string, when it is at

$60^{\circ}$ with vertical position.
(Mass of ball

$= 5\ kg$ ).

0%
$160\ N$
0%
$180\ N$
0%
$200\ N$
0%
$225\ N$

The position time plot for a

$400 gm$ object are shown. The impulses at

$0$ sec,

$1$ sec,

$3$ sec are :-

0%
$1.6 N-s, 1.6 N-s, 1.6 N-s$
0%
$1.6 N-s, 1.6 N-s, -1.6 N-s$
0%
$1.6 N-s, 0, 1.6 N-s$
0%
$1.6 N-s, 0, -1.6 N-s$

A mass of

$6 \ kg$ is suspended by a rope of length

$2 \ m$ from a ceiling. A force of

$50 \ N$ is applied in horizontal direction at the mid point of the rope. What is the angle of the rope with the vertical in equilibrium?

0%
${\tan ^{ - 1}}\left( {\dfrac{4}{5}} \right)$
0%
${\tan ^{ - 1}}\left( {\dfrac{5}{4}} \right)$
0%
${\tan ^{ - 1}}\left( {\dfrac{5}{6}} \right)$
0%
None of these

Explanation

Conditions of total equilibrium quint the sum of all

the external forces acting on the body is zero, and

the sum of all the torques from external forces

$c_{0}$

zero - These two Conditions must be simultaneously satisfied in equilibrium

for horizontal equilibrium

$T_{3}=T_{1} \sin \theta$

$for\ vertical\ equilibrium,$

$T_{2}=T_{1} \cos \theta$

$Taking\ ratios$

$\dfrac{T_{3}}{T_{2}}=\tan \theta$

for the equilibrium body

$T_{2}=m g$

$T_{2}=6 \times 10N \\$

$\therefore \quad \tan \theta=\dfrac{5}{6}$

$\theta_{2}$

$\theta=\tan ^{-1}\left(\dfrac{5}{6}\right)$

$\operatorname{option\ c}= \tan ^{-1}\left(\dfrac{5}{6}\right)$

A force time graph for the motion of a body is shown in figure. Change in linear momentum between

$0$ and

$8s$ is :

0%
$0$
0%
$4Ns$
0%
$8Ns$
0%
none

Explanation

The area under the graph gives linear momentum

Linear momentum=Force×time

Change in momentum=

$(-2 \times 2)+(4\times 1)+(-1\times 1)+(1\times 1)=5-5=0$

Change in momentum=0

CBSE Questions for Class 11 Engineering Physics Laws Of Motion Quiz 12 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Laws Of Motion Quiz 12 - MCQExams.com

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

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