Q501
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6. INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
In the arrangement shown in figure, the end A of light inextensible string is pulled with constant velocity $v = 6 \, ms^{-1}$ . The velocity of block B is (in $ms^{-1}$ )
Correct Answer: 2
Explanation:
According to Principle of Virtual Work, we have
$\begin{array}{r l} & T x = 3 T x _ {B} \\ \Rightarrow & x = 3 x _ {B} \end{array}$
$\Rightarrow \frac {d x}{d t} = 3 \frac {d x _ {B}}{d t}$
$\Rightarrow v = 3 v _ {B}$
$\Rightarrow v _ {B} = \frac {v}{3}$
$\begin{array}{r l} & T x = 3 T x _ {B} \\ \Rightarrow & x = 3 x _ {B} \end{array}$
$\Rightarrow \frac {d x}{d t} = 3 \frac {d x _ {B}}{d t}$
$\Rightarrow v = 3 v _ {B}$
$\Rightarrow v _ {B} = \frac {v}{3}$
Q502
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6. INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
Figure shows a string passing through two fixed pulley $P_{1}$ and $P_{3}$ and a pulley $P_{2}$ free to move vertically. One end of string is attached with ring A. Calculate the velocity of pulley $P_{2}$ at the instant shown in ms $^{-1}$ .
Correct Answer: 4
Explanation:
By constraint relation knowledge, we have
$\begin{array}{l l} & V _ {A} \sin (6 0 ^ {\circ}) = V _ {P _ {2}} [ 1 + \cos (6 0 ^ {\circ}) ] + V _ {B} \\ \Rightarrow & V _ {P _ {2}} = 4 \mathrm{ms} ^ {- 1} \end{array}$
$\begin{array}{l l} & V _ {A} \sin (6 0 ^ {\circ}) = V _ {P _ {2}} [ 1 + \cos (6 0 ^ {\circ}) ] + V _ {B} \\ \Rightarrow & V _ {P _ {2}} = 4 \mathrm{ms} ^ {- 1} \end{array}$
Q503
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6. INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
Find the tension in string 1. All the surfaces are frictionless. Strings are light and frictionless. (Take $g = 10 \, \mathrm{ms}^{-2}$ ).
Correct Answer: 8
Explanation:
$a = \frac {2 \times 1 0}{2 + \frac {1}{3} + 1} = 6 \mathrm{ms} ^ {- 2}$
Also, $2 \times g - T = 2 \times 6$
$\Rightarrow T = 8 \mathrm{N}$
Also, $2 \times g - T = 2 \times 6$
$\Rightarrow T = 8 \mathrm{N}$
Q504
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6. INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
A block of mass 1 kg is just fit in a groove in a platform kept horizontally. The groove is along +ve x-axis. The platform is given an acceleration $\vec{a}=2\hat{i}+3\hat{j}\ ms^{-2}$ . If block is not slipping on platform, then find the friction force acting on block (in newton).
Correct Answer: 2
Explanation:
Since the block is not moving, so frictional force acting on the block is opposite to $\vec{F}_{\text{pseudo}} = -(ma)\hat{i}$
$\Rightarrow \vec {f} = - \vec {F} _ {\text { pseudo }} = (2 \hat {i}) \text { newton }$
$\Rightarrow \vec {f} = - \vec {F} _ {\text { pseudo }} = (2 \hat {i}) \text { newton }$
Q505
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6. INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
Calculate the magnitude of the force exerted by A on B.
All surfaces are frictionless.
All surfaces are frictionless.
Correct Answer: 8
Explanation:
$a = \frac{10}{5} = 2\mathrm{ms}^{-2}$
$f = \left(M _ {B} + M _ {C}\right) \times a = 8 \mathrm{N}$
$f \rightarrow \boxed {B} \boxed {C}$
$f = \left(M _ {B} + M _ {C}\right) \times a = 8 \mathrm{N}$
$f \rightarrow \boxed {B} \boxed {C}$
Q506
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6. INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
Two blocks 1 and 2 of mass 2 kg and 4 kg are kept connected as shown in figure. All pulley and string are massless and surfaces are frictionless. Calculate the acceleration of block 1 at the instant shown.
Correct Answer: 0
Explanation:
Free body diagram for the middle pulley is shown.
So, we observe that
$2T-T=0$
($\because$ pulley is light, so $m_{\text{pulley}}\approx0$)
$\Rightarrow T=0$
So, we observe that
$2T-T=0$
($\because$ pulley is light, so $m_{\text{pulley}}\approx0$)
$\Rightarrow T=0$
Q507
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6. INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
Two blocks A and B of equal mass m are connected through a massless string and arranged as shown in figure. Friction is absent everywhere. When the system is released from rest, then calculate the tension in string. Where m is 2 kg. $g = 10 \, \text{ms}^{-2}$
Correct Answer: 5
Explanation:
For B, $mg-T=ma$ ā¦ (1)
For A, $T+mg\sin\theta=ma$ ā¦ (2)
Adding (1) and (2), we get
$a=\frac{g}{2}(1+\sin\theta)$
$\Rightarrow a=\frac{3g}{4}$
Put $a=\frac{3g}{4}$ in (1), we get
$mg-T=\frac{3mg}{4}$
$\Rightarrow T=\frac{mg}{4}=5\ \text{N}$
For A, $T+mg\sin\theta=ma$ ā¦ (2)
Adding (1) and (2), we get
$a=\frac{g}{2}(1+\sin\theta)$
$\Rightarrow a=\frac{3g}{4}$
Put $a=\frac{3g}{4}$ in (1), we get
$mg-T=\frac{3mg}{4}$
$\Rightarrow T=\frac{mg}{4}=5\ \text{N}$
Q508
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6. INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
Assuming all the surfaces to be frictionless, find the acceleration of the block C in $ms^{-2}$ shown in figure.
Correct Answer: 2
Explanation:
$a_C=\frac{a_A+a_B}{2}$
$\Rightarrow a_C=\frac{7-3}{2}=2\ \text{m s}^{-2}$
$\Rightarrow a_C=\frac{7-3}{2}=2\ \text{m s}^{-2}$
Q509
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6. INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
A block of mass 2 kg is placed on rough horizontal surface (coefficient of friction = 0.2) and is pulled by horizontal force $F = (2t)$ N where t is time in sec. Find the velocity of block at t = 4 seconds.
Correct Answer: 4
Explanation:
$f_l=(0.2)(20)=4\ \text{N}$
So, till $2t=4\Rightarrow t=2\ \text{s}$, block will be at rest.
For $2\leq t\leq4\ \text{s}$, the block will move with an acceleration
$a=\frac{F-f_l}{m}$
$\Rightarrow \frac{dv}{dt}=(2t-4)$
$\Rightarrow \int_0^v dv=2\int_2^4t,dt-4\int_2^4dt$
$\Rightarrow v=2\left[\frac{t^2}{2}\right]_2^4-4[t]_2^4$
$\Rightarrow v=(16-4)-4(4-2)$
$\Rightarrow v=12-8$
$\Rightarrow v=4\ \text{m s}^{-1}$
So, till $2t=4\Rightarrow t=2\ \text{s}$, block will be at rest.
For $2\leq t\leq4\ \text{s}$, the block will move with an acceleration
$a=\frac{F-f_l}{m}$
$\Rightarrow \frac{dv}{dt}=(2t-4)$
$\Rightarrow \int_0^v dv=2\int_2^4t,dt-4\int_2^4dt$
$\Rightarrow v=2\left[\frac{t^2}{2}\right]_2^4-4[t]_2^4$
$\Rightarrow v=(16-4)-4(4-2)$
$\Rightarrow v=12-8$
$\Rightarrow v=4\ \text{m s}^{-1}$
Q510
Advance
6. INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
In the arrangement shown in figure, the mass of the rod M = 2 kg exceeds the mass m = 0.5 kg of the ball. The mass has an opening permitting it to slide along the thread with some friction. The mass of the pulley and the friction in its axle are negligible. At the initial moment ball was located opposite to the lower end of the rod. When set free, both bodies began moving with constant accelerations. Find the friction force between the ball and the thread if t = 2 second after the beginning of motion of the ball, the ball got opposite the
upper end of the rod. The rod length equals l = 0.3 m. (in $10^{-1}$ ).
Correct Answer: 1
Explanation:
For rod of mass $2\ \text{kg}$,
$20-T=2a_1$ ā¦ (1)
For ball,
$5-f=5a_2$ ā¦ (2)
Since thread is mass less,
$T=f$ ā¦ (3)
$l=0.3=\frac{1}{2}a_{\text{rel}}t^2$
$20-T=2a_1$ ā¦ (1)
For ball,
$5-f=5a_2$ ā¦ (2)
Since thread is mass less,
$T=f$ ā¦ (3)
$l=0.3=\frac{1}{2}a_{\text{rel}}t^2$
Q511
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6. INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
Three blocks A, B and C of mass m each are arranged in pulley mass system as shown. Coefficient of friction between block A and horizontal surface is equal to 0.5 and a force P acts on A in the direction shown. Calculate $\frac{P}{mg}$ so that block 'C' doesn't move.
Correct Answer: 5
Explanation:
If $C$ doesn't move, then
$a_A=4a_B$ ā¦ (1)
For A,
$P-T-\mu mg=m(4a_B)$ ā¦ (2)
For B,
$4T-mg=ma_B$ ā¦ (3)
$\Rightarrow \frac{P}{mg}=5$
$a_A=4a_B$ ā¦ (1)
For A,
$P-T-\mu mg=m(4a_B)$ ā¦ (2)
For B,
$4T-mg=ma_B$ ā¦ (3)
$\Rightarrow \frac{P}{mg}=5$
Q512
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7. ARCHIVE: JEE MAIN
MCQ
A particle of mass $m$ is at rest at the origin at time $t = 0$ . It is subjected to a force $F(t) = F_0 e^{-bt}$ in the $x$ direction. Its speed $v(t)$ is depicted by which of the following curves?
A.
B.
C.
D.
Q513
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7. ARCHIVE: JEE MAIN
MCQ
The figure shows the position-time $(x - t)$ graph of one-dimensional motion of a body of mass $0.4\mathrm{kg}$ . The magnitude of each impulse is
A.
0.2 Ns
B.
0.4 Ns
C.
0.8 Ns
D.
1.6 Ns