JEE (Main) 2022
Q1
Allen
JEE Main PYQ
Numerical
A system to 10 balls each of mass 2 kg are connected via massless and unstretchable string. The system is allowed to slip over the edge of a smooth table as shown in figure. Tension on the string between the $7^{th}$ and $8^{th}$ ball is ____ N when $6^{th}$ ball just leaves the table.
Correct Answer: 36
Explanation:
Ans. (36)
$a = \frac {6 m g}{1 0 m} = \frac {6 g}{1 0} = \frac {3 g}{5}$
Taking 8, 9, 10 together
$\begin{array}{r l} & T = 3 m a \\ & = 3 m \times \frac {3 g}{5} \\ & = 3 6 N \end{array}$
$a = \frac {6 m g}{1 0 m} = \frac {6 g}{1 0} = \frac {3 g}{5}$
Taking 8, 9, 10 together
$\begin{array}{r l} & T = 3 m a \\ & = 3 m \times \frac {3 g}{5} \\ & = 3 6 N \end{array}$
JEE (Advanced) 2022
Q2
Errorless
JEE Section
Numerical
A particle of mass 1 kg is subjected to a force which depends on the position as $\vec{F} = -k(x\hat{i} + y\hat{j})kgms^{-2}$ with $k = 1kgs^{-2}$ . At time $t = 0$ , the particle's position $\vec{r} = \left(\frac{1}{\sqrt{2}}\hat{i} + \sqrt{2}\hat{j}\right)m$ and its velocity $\vec{v} = \left(-\sqrt{2}\hat{i} + \sqrt{2}\hat{j} + \frac{2}{\pi}\hat{k}\right)ms^{-1}$ . Let $v_x$ and $v_y$ denote the $x$ and the $y$ components of the particle's velocity respectively. Ignore gravity when $z = 0.5m$ , the value of $(xv_y - yv_x)$ is ____ $m^2 s^{-1}$
Correct Answer: 3
Explanation:
$F_{x} = -x = ma_{x}$
So $a_{x} = \frac{d^{2}x}{dt^{2}} = -x$
$\Rightarrow x = A _ {x} \sin (\omega t + \phi_ {x}) \quad (\omega = 1 r a d / s)$
and $v_{x} = A_{x}\omega \cos (\omega t + \phi_{x})$
at $t = 0$ , $x = \frac{1}{\sqrt{2}} m$ and $v_{x} = -\sqrt{2} m / s$
$\text { So } A _ {x} \sin \phi_ {x} = \frac {1}{\sqrt {2}} \text { and } A _ {x} \cos \phi_ {x} = - \sqrt {2}$
$\Rightarrow \tan \phi_ {x} = - \frac {1}{2}\tag{... (i}$
and $A_{x}=\sqrt{\frac{5}{2}}m$
... (ii)
Similarly, $F_{y} = -y = ma_{y}$
$\Rightarrow \frac {d ^ {2} y}{d t ^ {2}} = - y$
$\mathrm{So}, y = A _ {y} \sin (\omega t + \phi_ {y}) \quad (\omega = 1 r a d / s)$
and $v_{y} = A_{y}\omega \cos (\omega t + \phi_{y})$
at $t = 0, y = \sqrt{2} m$ and $v_{y} = \sqrt{2} m / s$
So $A_{y}\sin \phi_{y} = \sqrt{2}$ and $A_{y}\cos \phi_{y} = \sqrt{2}$
$\Rightarrow \phi = \frac {\pi}{4}\tag{... (iii}$
and $A_{y} = 2$
$\begin{array}{r l} & {\mathrm{So}, (x v _ {y} - y v _ {x}) = \sqrt {\frac {5}{2}} \sin (\omega t + \phi_ {x}) \times 2 \cos (\omega t + \phi_ {y}) -} \\ & {\qquad \qquad \qquad \qquad \qquad 2 \sin (\omega t + \phi_ {y}) \times \sqrt {\frac {5}{2}} \cos (\omega t + \phi_ {x})} \\ & {\Rightarrow x v _ {y} - y v _ {x} = \sqrt {\frac {5}{2}} \times 2 [ \sin (\omega t + \phi_ {x}) \cos (\omega t + \phi_ {y}) -} \\ & {\qquad \qquad \qquad \qquad \qquad \sin (\omega t + \phi_ {y}) \cos (\omega t + \phi_ {x}) ]} \\ & {\Rightarrow x v _ {y} - y v _ {x} = \sqrt {1 0} \sin (\phi_ {x} - \phi_ {y})} \\ & {\Rightarrow x v _ {y} - y v _ {x} = \sqrt {1 0} [ \sin \phi_ {x} \cos \phi_ {y} - \cos \phi_ {x} \sin \phi_ {y} ]} \\ & {\Rightarrow x v _ {y} - y v _ {x} = \sqrt {1 0} \left[ \frac {1}{\sqrt {5}} \times \frac {1}{\sqrt {2}} - \left(- \frac {2}{\sqrt {5}}\right) \times \frac {1}{\sqrt {2}} \right]} \\ & {\Rightarrow x v _ {y} - y v _ {x} = 3.} \end{array}\tag{... (iv}$
So $a_{x} = \frac{d^{2}x}{dt^{2}} = -x$
$\Rightarrow x = A _ {x} \sin (\omega t + \phi_ {x}) \quad (\omega = 1 r a d / s)$
and $v_{x} = A_{x}\omega \cos (\omega t + \phi_{x})$
at $t = 0$ , $x = \frac{1}{\sqrt{2}} m$ and $v_{x} = -\sqrt{2} m / s$
$\text { So } A _ {x} \sin \phi_ {x} = \frac {1}{\sqrt {2}} \text { and } A _ {x} \cos \phi_ {x} = - \sqrt {2}$
$\Rightarrow \tan \phi_ {x} = - \frac {1}{2}\tag{... (i}$
and $A_{x}=\sqrt{\frac{5}{2}}m$
... (ii)
Similarly, $F_{y} = -y = ma_{y}$
$\Rightarrow \frac {d ^ {2} y}{d t ^ {2}} = - y$
$\mathrm{So}, y = A _ {y} \sin (\omega t + \phi_ {y}) \quad (\omega = 1 r a d / s)$
and $v_{y} = A_{y}\omega \cos (\omega t + \phi_{y})$
at $t = 0, y = \sqrt{2} m$ and $v_{y} = \sqrt{2} m / s$
So $A_{y}\sin \phi_{y} = \sqrt{2}$ and $A_{y}\cos \phi_{y} = \sqrt{2}$
$\Rightarrow \phi = \frac {\pi}{4}\tag{... (iii}$
and $A_{y} = 2$
$\begin{array}{r l} & {\mathrm{So}, (x v _ {y} - y v _ {x}) = \sqrt {\frac {5}{2}} \sin (\omega t + \phi_ {x}) \times 2 \cos (\omega t + \phi_ {y}) -} \\ & {\qquad \qquad \qquad \qquad \qquad 2 \sin (\omega t + \phi_ {y}) \times \sqrt {\frac {5}{2}} \cos (\omega t + \phi_ {x})} \\ & {\Rightarrow x v _ {y} - y v _ {x} = \sqrt {\frac {5}{2}} \times 2 [ \sin (\omega t + \phi_ {x}) \cos (\omega t + \phi_ {y}) -} \\ & {\qquad \qquad \qquad \qquad \qquad \sin (\omega t + \phi_ {y}) \cos (\omega t + \phi_ {x}) ]} \\ & {\Rightarrow x v _ {y} - y v _ {x} = \sqrt {1 0} \sin (\phi_ {x} - \phi_ {y})} \\ & {\Rightarrow x v _ {y} - y v _ {x} = \sqrt {1 0} [ \sin \phi_ {x} \cos \phi_ {y} - \cos \phi_ {x} \sin \phi_ {y} ]} \\ & {\Rightarrow x v _ {y} - y v _ {x} = \sqrt {1 0} \left[ \frac {1}{\sqrt {5}} \times \frac {1}{\sqrt {2}} - \left(- \frac {2}{\sqrt {5}}\right) \times \frac {1}{\sqrt {2}} \right]} \\ & {\Rightarrow x v _ {y} - y v _ {x} = 3.} \end{array}\tag{... (iv}$
JEE (Main) 2021
Q3
Allen
JEE Main PYQ
Numerical
The coefficient of static friction between a wooden block of mass 0.5 kg and a vertical rough wall is 0.2. The magnitude of horizontal force that should be applied on the block to keep it adhere to the wall will be ____ N. $[g = 10 \, ms^{-2}]$
Correct Answer: 25
Explanation:
Ans. (25)
F.B.D. of the block is shown in the diagram Since block is at rest therefore
f r - m g = 0
$F - N = 0\tag{... (1}$
$f r \leq \mu N\tag{... (2}$
In limiting case
$f r = \mu N = \mu F\tag{... (3}$
Using equation (1) and (3)
$\therefore \mu F = m g$
$\Rightarrow F = \frac {0 . 5 \times 1 0}{0 . 2} = 2 5 N$
F.B.D. of the block is shown in the diagram Since block is at rest therefore
f r - m g = 0
$F - N = 0\tag{... (1}$
$f r \leq \mu N\tag{... (2}$
In limiting case
$f r = \mu N = \mu F\tag{... (3}$
Using equation (1) and (3)
$\therefore \mu F = m g$
$\Rightarrow F = \frac {0 . 5 \times 1 0}{0 . 2} = 2 5 N$
JEE (Main) 2021
Q4
Allen
JEE Main PYQ
Numerical
As shown in the figure, a block of mass $\sqrt{3}$ kg is kept on a horizontal rough surface of coefficient of friction $\frac{1}{3\sqrt{3}}$ . The critical force to be applied on the vertical surface as shown at an angle $60^{\circ}$ with horizontal such that it does not move, will be 3x. The value of x will be
$[g = 10m / s^2;\sin 60^\circ = \frac{\sqrt{3}}{2};\cos 60^\circ = \frac{1}{2} ]$
Correct Answer: 3
Explanation:
Ans. (3)
$F\cos 60^{\circ} = \mu N$ or $\frac{F}{2} = \frac{1}{3\sqrt{3}} N$
...(1)
and $N = \sin 60^{\circ} + \sqrt{3} g$
...(2)
From equation 1) and 2)
${\frac {F}{2}} = {\frac {1}{3 \sqrt {3}}} \left({\frac {F \sqrt {3}}{2}} + \sqrt {3} g\right)$
$\Rightarrow \quad F = g = 1 0 \mathrm{Newton} = 3 x$
So $x = \frac{10}{3} = 3.33$
$F\cos 60^{\circ} = \mu N$ or $\frac{F}{2} = \frac{1}{3\sqrt{3}} N$
...(1)
and $N = \sin 60^{\circ} + \sqrt{3} g$
...(2)
From equation 1) and 2)
${\frac {F}{2}} = {\frac {1}{3 \sqrt {3}}} \left({\frac {F \sqrt {3}}{2}} + \sqrt {3} g\right)$
$\Rightarrow \quad F = g = 1 0 \mathrm{Newton} = 3 x$
So $x = \frac{10}{3} = 3.33$
JEE (Main) 2021
Q5
Allen
JEE Main PYQ
Numerical
A boy of mass 4 kg is standing on a piece of wood having mass 5 kg. If the coefficient of friction between the wood and the floor is 0.5, the maximum force that the boy can exert on the rope so that the piece of wood does not move from its place is ____N. (Round off to the Nearest Integer) [Take $g = 10ms^{-2}$ ]
Correct Answer: 30
Explanation:
Ans. (30)
$N + T = 9 0$
$T = \mu N = 0.5 (9 0 - T)$
$1.5 T = 4 5$
$T = 3 0$
$N + T = 9 0$
$T = \mu N = 0.5 (9 0 - T)$
$1.5 T = 4 5$
$T = 3 0$
JEE (Main) 2021
Q6
Allen
JEE Main PYQ
Numerical
When a body slides down from rest along a smooth inclined plane making an angle of $30^{\circ}$ with the horizontal, it takes time T. When the same body slides down from the rest along a rough inclined plane making the same angle and through the same distance, it takes time $\alpha T$ , where is $\alpha$ constant greater than 1. The co-efficient of friction between the body and the rough plane is $\frac{1}{\sqrt{x}}\left(\frac{\alpha^{2}-1}{\alpha^{2}}\right)$ where x = .........
Correct Answer: 3
Explanation:
Ans. (3)
On smooth incline
$a = g \sin 3 0 ^ {\circ}$
By $S = ut + \frac{1}{2} at^{2}$
$S = \frac {1}{2} \frac {g}{2} T ^ {2} = \frac {g}{4} T ^ {2}\tag{... (i}$
On rough incline
$a = g \sin 3 0 ^ {\circ} - \mu g \cos 3 0 ^ {\circ}$
By $S = ut + \frac{1}{2} at^{2}$
$S = \frac {1}{4} g (1 - \sqrt {3} \mu) (\alpha T) ^ {2}\tag{... (ii}$
By (i) and (ii)
$\frac {1}{4} g T ^ {2} = \frac {1}{4} g (1 - \sqrt {3} \mu) \alpha^ {2} T ^ {2}$
$\begin{array}{r l} \Rightarrow & 1 - \sqrt {3} \mu = \frac {1}{\alpha^ {2}} \\ \Rightarrow & \mu = \left(\frac {\alpha^ {2} - 1}{\alpha^ {2}}\right) \cdot \frac {1}{\sqrt {3}} \\ \Rightarrow & x = 3. 0 0 \end{array}$
On smooth incline
$a = g \sin 3 0 ^ {\circ}$
By $S = ut + \frac{1}{2} at^{2}$
$S = \frac {1}{2} \frac {g}{2} T ^ {2} = \frac {g}{4} T ^ {2}\tag{... (i}$
On rough incline
$a = g \sin 3 0 ^ {\circ} - \mu g \cos 3 0 ^ {\circ}$
By $S = ut + \frac{1}{2} at^{2}$
$S = \frac {1}{4} g (1 - \sqrt {3} \mu) (\alpha T) ^ {2}\tag{... (ii}$
By (i) and (ii)
$\frac {1}{4} g T ^ {2} = \frac {1}{4} g (1 - \sqrt {3} \mu) \alpha^ {2} T ^ {2}$
$\begin{array}{r l} \Rightarrow & 1 - \sqrt {3} \mu = \frac {1}{\alpha^ {2}} \\ \Rightarrow & \mu = \left(\frac {\alpha^ {2} - 1}{\alpha^ {2}}\right) \cdot \frac {1}{\sqrt {3}} \\ \Rightarrow & x = 3. 0 0 \end{array}$
JEE (Main) 2020
Q7
Allen
JEE Main PYQ
Numerical
A small block starts slipping down from a point B on an inclined plane AB, which is making an angle $\theta$ with the horizontal section BC is smooth and the remaining section CA is rough with a coefficient of friction $\mu$ . It is found that the block comes to rest as it reaches the bottom (point A) of the inclined plane. If BC = 2AC, the coefficient of friction is given by $\mu = k \tan\theta$ . The value of k is ____.
Correct Answer: 3
Explanation:
Ans. (3)
Apply work energy theorem
$mg \sin \theta (AC + 2AC) - \mu mg \cos \theta AC = 0$
$\mu = 3 \tan \theta$
Apply work energy theorem
$mg \sin \theta (AC + 2AC) - \mu mg \cos \theta AC = 0$
$\mu = 3 \tan \theta$
JEE (Advanced) 2020
Q8
Errorless
Integer Type Questions-friction
Numerical
Put a uniform meter scale horizontally on your extended index fingers with the left one at 0.00 cm and the right one at 90.00 cm. When you attempt to move both the fingers slowly towards the center, initially only the left finger slips with respect to the scale and the right finger does not. After some distance, the left finger stops and the right one starts slipping. Then the right finger stops at a distance $x_{R}$ from the center (50.00 cm) of the scale and the left one starts slipping again. This happens because of the difference in the frictional forces on the two fingers. If the coefficients of static and dynamic friction between the fingers and the scale are 0.40 and 0.32, respectively, the value of $x_{R}$ (in cm) is ____
Correct Answer: B
JEE (Advanced) 2018
Q9
Errorless
JEE Section
Numerical
A solid horizontal surface is covered with a thin layer of oil. A rectangular block of mass $m = 0.4\mathrm{kg}$ is at rest on this surface. An impulse of $1.0N s$ is applied to the block at time to $t = 0$ so that it starts moving along the x-axis with a velocity $v(t) = v_0e^{-t / \tau}$ , where $v_0$ is a constant and $\tau = 4s$ . The displacement of the block, in metres, at $t = \tau$ is ____ Take $e^{-1} = 0.37$
Correct Answer: 6.30
Explanation:
$J = 1 \longrightarrow \boxed {m = 0. 4}$
$v = v _ {0} e ^ {- t / \tau}$
$v _ {0} = \frac {J}{m} = 2. 5 m / s$
$v = v _ {0} e ^ {- t / \tau} \Rightarrow \frac {d x}{d t} = v _ {0} e ^ {- t / \tau}$
$\int_ {0} ^ {x} d x = v _ {0} \int_ {0} ^ {\tau} e ^ {- t / \tau} d t$
$\int e ^ {- x} d x = \frac {e ^ {- x}}{- 1}$
$x = v _ {0} \left[ \frac {e ^ {- t / \tau}}{- \frac {1}{\tau}} \right] _ {0} ^ {\tau}$
$x = 2.5 (- 4) (e ^ {- 1} - e ^ {0}) \Rightarrow x = 6.3 0.$
$v = v _ {0} e ^ {- t / \tau}$
$v _ {0} = \frac {J}{m} = 2. 5 m / s$
$v = v _ {0} e ^ {- t / \tau} \Rightarrow \frac {d x}{d t} = v _ {0} e ^ {- t / \tau}$
$\int_ {0} ^ {x} d x = v _ {0} \int_ {0} ^ {\tau} e ^ {- t / \tau} d t$
$\int e ^ {- x} d x = \frac {e ^ {- x}}{- 1}$
$x = v _ {0} \left[ \frac {e ^ {- t / \tau}}{- \frac {1}{\tau}} \right] _ {0} ^ {\tau}$
$x = 2.5 (- 4) (e ^ {- 1} - e ^ {0}) \Rightarrow x = 6.3 0.$
JEE (Advanced) 2018
Q10
Advance
ARCHIVE: JEE ADVANCED
Numerical
A solid horizontal surface is covered with a thin layer of oil. A rectangular block of mass $m = 0.4 \, \mathrm{kg}$ is at rest on this surface. An impulse of 1 Ns is applied to the block at time $t = 0$ , so that it starts moving along the X-axis with a velocity $v(t) = v_0 e^{-t / \tau}$ , where $v_0$ is a constant and $\tau = 4 \, \mathrm{s}$ . The displacement of the block, in metres, at $t = \tau$ is ...... (Take, $e^{-1} = 0.37$ ) [Round off your answer to nearest integer].
Correct Answer: 6
Explanation:
Linear impulse, $J = mv_{0}$
$\Rightarrow v _ {0} = \frac {J}{m} = 2. 5 \mathrm{ms} ^ {- 1}$
$\Rightarrow v = v _ {0} e ^ {- t / \tau}$
$\Rightarrow \frac {d x}{d t} = v _ {0} e ^ {- t / \tau}$
$\Rightarrow \int_ {0} ^ {x} d x = v _ {0} \int_ {0} ^ {\tau} e ^ {- t / \tau} d t$
$\Rightarrow x = v _ {0} \left[ \frac {e ^ {- t / \tau}}{- 1 / \tau} \right] _ {0} ^ {\tau}$
$\Rightarrow x = 2. 5 (- 4) \big (e ^ {- 1} - e ^ {0} \big)$
$\Rightarrow x = 2. 5 (- 4) (0. 3 7 - 1)$
$\Rightarrow x = 6. 3 0 \mathrm{m}$
$\Rightarrow x = 6 \mathrm{m}$
$\Rightarrow v _ {0} = \frac {J}{m} = 2. 5 \mathrm{ms} ^ {- 1}$
$\Rightarrow v = v _ {0} e ^ {- t / \tau}$
$\Rightarrow \frac {d x}{d t} = v _ {0} e ^ {- t / \tau}$
$\Rightarrow \int_ {0} ^ {x} d x = v _ {0} \int_ {0} ^ {\tau} e ^ {- t / \tau} d t$
$\Rightarrow x = v _ {0} \left[ \frac {e ^ {- t / \tau}}{- 1 / \tau} \right] _ {0} ^ {\tau}$
$\Rightarrow x = 2. 5 (- 4) \big (e ^ {- 1} - e ^ {0} \big)$
$\Rightarrow x = 2. 5 (- 4) (0. 3 7 - 1)$
$\Rightarrow x = 6. 3 0 \mathrm{m}$
$\Rightarrow x = 6 \mathrm{m}$
JEE (Advanced) 2014
Q11
Errorless
Matrix Match Type Questions
Numerical
A person in a lift is holding a water jar, which has a small hole at the lower end of its side. When the lift is at rest, the water jet coming out of the hole hits the floor of the lift at a distance d of 1.2 m from the person. In the following, state of the lift's motion is given in Column-I and the distance where the water jet hits the floor of the lift is given in Column-II. Match the statements from Column-I with those in Column-II and select the correct answer using the code given below the columns
| Column-I | Column-II | ||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
|
|
Correct Answer: A ā p; B ā p; C ā p; D ā s
Explanation:
$\mathbf{A} \rightarrow \mathbf{p}; \mathbf{B} \rightarrow \mathbf{p}; \mathbf{C} \rightarrow \mathbf{p}; \mathbf{D} \rightarrow \mathbf{s}$
Match the column
When lift is at rest:
$d = \sqrt {4 h _ {1} h _ {2}}$
A) $g_{eff} > g$ $d = \sqrt{4h_1h_2} = 1.2m$
B) $g_{eff} < g$ $d = \sqrt{4h_1h_2} = 1.2m$
C) $g_{eff} = g$ $d = \sqrt{4h_1h_2} = 1.2m$
D) $g_{eff} = 0$ No water leaks out of the jar.
Match the column
When lift is at rest:
$d = \sqrt {4 h _ {1} h _ {2}}$
A) $g_{eff} > g$ $d = \sqrt{4h_1h_2} = 1.2m$
B) $g_{eff} < g$ $d = \sqrt{4h_1h_2} = 1.2m$
C) $g_{eff} = g$ $d = \sqrt{4h_1h_2} = 1.2m$
D) $g_{eff} = 0$ No water leaks out of the jar.
JEE (Advanced) 2014
Q12
Errorless
Matrix Match Type Questions-friction
Numerical
A block of mass $m_{1}=1kg$ another mass $m_{2}=2kg$ , are placed together (see figure) on an inclined plane with angle of inclination $\theta$ . Various values of $\theta$ are given in Column-I. The coefficient of friction between the block $m_{1}$ and the Plane is always zero. The coefficient of static and dynamic friction between the block $m_{2}$ and the plane are equal to $\mu=0.3$ . In Column-II expression for the friction on block $m_{2}$ given. Match the correct expression of the friction in Column-II with the angles given in Column-I, and choose the correct option. The acceleration due to gravity is denoted by g
[Useful information : $\tan(5.5^{\circ})\approx0.1;\tan(11.5^{\circ})\approx0.2;$
$\tan (16.5^{\circ}\approx 0.3)]$
[Useful information : $\tan(5.5^{\circ})\approx0.1;\tan(11.5^{\circ})\approx0.2;$
$\tan (16.5^{\circ}\approx 0.3)]$
| Column-I | Column-II | ||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
|
|
Correct Answer: 6.30
Explanation:
$\mathbf{A}\rightarrow \mathbf{q};\mathbf{B}\rightarrow \mathbf{q};\mathbf{C}\rightarrow \mathbf{r};\mathbf{D}\rightarrow \mathbf{r}$
Block will not slip if
$(m_{1} + m_{2})g\sin \theta \leq \mu m_{2}g\cos \theta$
$3\sin \theta \leq \left(\frac{3}{10}\right)(2)\cos \theta$
$\tan \theta \leq \frac {1}{5} \Rightarrow \theta \leq 11.5 ^ {\circ}$
A) $\theta = 5^{\circ}$ friction is static $f = (m_{1} + m_{2})g\sin \theta$
B) $\theta = 10^{\circ}$ friction is static $f = (m_{1} + m_{2})g\sin \theta$
C) $\theta = 15^{\circ}$ friction is kinetic $f = \mu m_{2}g\cos \theta$
D) $\theta = 20^{\circ}$ friction is kinetic $f = \mu m_{2}g\cos \theta$
Block will not slip if
$(m_{1} + m_{2})g\sin \theta \leq \mu m_{2}g\cos \theta$
$3\sin \theta \leq \left(\frac{3}{10}\right)(2)\cos \theta$
$\tan \theta \leq \frac {1}{5} \Rightarrow \theta \leq 11.5 ^ {\circ}$
A) $\theta = 5^{\circ}$ friction is static $f = (m_{1} + m_{2})g\sin \theta$
B) $\theta = 10^{\circ}$ friction is static $f = (m_{1} + m_{2})g\sin \theta$
C) $\theta = 15^{\circ}$ friction is kinetic $f = \mu m_{2}g\cos \theta$
D) $\theta = 20^{\circ}$ friction is kinetic $f = \mu m_{2}g\cos \theta$
IIT-JEE 2011
Q13
Errorless
Integer Type Questions-friction
Numerical
A block is moving on an inclined plane making an angle $45^{\circ}$ with the horizontal and the coefficient of friction is $\mu$ . The force required to just push it up the inclined plane is 3 times the force required to just prevent it from sliding down. If we define $N = 10\mu$ , then N is
Correct Answer: C
IIT-JEE 2011
Q14
Advance
ARCHIVE: JEE ADVANCED
Numerical
A block is moving on an inclined plane making an angle $45^{\circ}$ with the horizontal and the coefficient of friction is $\mu$ . The force required to just push it up the inclined plane is 3 times the force required to just prevent it from sliding down. If define $N = 10 \mu$ , then N is
Correct Answer: 5
Explanation:
$F _ {1} = m g \sin \theta + \mu m g \cos \theta$
$F _ {2} = m g \sin \theta - \mu m g \cos \theta$
Given that $F_{1}=3F_{2}$
$\Rightarrow \left(\sin 4 5 ^ {\circ} + \mu \cos 4 5 ^ {\circ}\right)$
$\Rightarrow F _ {1} = 3 \left(\sin 4 5 ^ {\circ} - \mu \cos 4 5 ^ {\circ}\right)$
On solving, we get
$\mu = 0.5$
$\Rightarrow N = 1 0 \mu = 5$
KCET 2009
Q15
Errorless
Equilibrium of Forces
Numerical
Three concurrent co-planar forces 1 N, 2 N and 3 N acting along different directions on a body
(a) Can keep the body in equilibrium if $2N$ and $3N$ act at right angle
(b) Can keep the body in equilibrium if $1N$ and $2N$ act at right angle
(d) Can keep the body in equilibrium if $1N$ and $3N$ act at an acute angle
Correct Answer: 3
Explanation:
If $1N$ and $2N$ act in the same direction and $3N$ acts in opposite direction, equilibrium is possible.
Q16
Allen
Newtons Laws of Motion
Numerical
Draw free body diagram of block of mass m.
(A)
(B)
(C)
(D)
(E)
(F)
(G)
(H)
(I)
(A)
(B)
(C)
(D)
(E)
(F)
(G)
(H)
(I)
Correct Answer: (a)
Explanation:
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
Q17
Allen
Newtons Laws of Motion
Numerical
In the given figure, what is the minimum possible tension $(N)$ in the string 2?
Correct Answer: 8
Explanation:
Ans. 8
$T_{1}\sin 37^{\circ} + T_{2}\cos \theta = 10$
$T_{1}\cos 37^{\circ} = T_{2}\sin \theta$
$\sin 37^{\circ}\left(\frac{T_{2}\sin\theta}{\cos 37^{\circ}}\right) + T_{2}\cos \theta = 10$
$T _ {2} = \frac {4 0}{3 \sin \theta + 4 \cos \theta}$
$T_{2\ min}$ occurs if $(3\sin\theta+4\cos\theta)$ is maximum
$T _ {2 \min} = \frac {4 0}{5} = 8$
$T_{1}\sin 37^{\circ} + T_{2}\cos \theta = 10$
$T_{1}\cos 37^{\circ} = T_{2}\sin \theta$
$\sin 37^{\circ}\left(\frac{T_{2}\sin\theta}{\cos 37^{\circ}}\right) + T_{2}\cos \theta = 10$
$T _ {2} = \frac {4 0}{3 \sin \theta + 4 \cos \theta}$
$T_{2\ min}$ occurs if $(3\sin\theta+4\cos\theta)$ is maximum
$T _ {2 \min} = \frac {4 0}{5} = 8$
Q18
Allen
Newtons Laws of Motion
Numerical
A ladder is hanging from ceiling as shown in figure. Three men A, B and C of masses 40 kg, 60 kg, and 50 kg are climbing the ladder. Man A is going up with retardation $2 \, m/s^{2}$ , C is going up with an acceleration of $1 \, m/s^{2}$ and man B is going up with a constant speed of 0.5 m/s. Find the tension in the string supporting the ladder. $[g = 9.8 \, m/s^{2}]$
Correct Answer: $1440N$
Explanation:
Ans. 1440 N
For man B,
$T _ {B} - 6 0 \times 9. 8 - 5 4 0 = 0$
$T _ {B} = 5 4 0 + 5 8 8 = 1 1 2 8 N$
For man A,
$4 0 \mathrm{g} + 1 1 2 8 - T _ {A} = 4 0 \times 2$
$T _ {A} = 3 9 2 + 3 9 2 + 1 1 2 8 - 1 6 0$
$\boxed {T _ {A} = 1 4 4 0 N}$
For man C,
$T _ {C} - 5 0 g = 5 0 \times 1$
$T _ {C} = 5 0 g = (1 + g)$
$T _ {C} = 5 0 (1 + 9. 8)$
$T _ {C} = 1 0. 8 \times 5 0$
$T _ {C} = 4 5 0 N$
For man B,
$T _ {B} - 6 0 \times 9. 8 - 5 4 0 = 0$
$T _ {B} = 5 4 0 + 5 8 8 = 1 1 2 8 N$
For man A,
$4 0 \mathrm{g} + 1 1 2 8 - T _ {A} = 4 0 \times 2$
$T _ {A} = 3 9 2 + 3 9 2 + 1 1 2 8 - 1 6 0$
$\boxed {T _ {A} = 1 4 4 0 N}$
For man C,
$T _ {C} - 5 0 g = 5 0 \times 1$
$T _ {C} = 5 0 g = (1 + g)$
$T _ {C} = 5 0 (1 + 9. 8)$
$T _ {C} = 1 0. 8 \times 5 0$
$T _ {C} = 4 5 0 N$
Q19
Allen
Newtons Laws of Motion
Numerical
A 40 kg monkey climbs a rope that passes over an ideal pulley. The other end of the rope is attached to a 60 kg weight placed on the ground. What is the maximum upward acceleration the monkey can have without lifting the weight? If he climbs the rope with upward acceleration 2g, with what acceleration the weight will rise up?
Correct Answer: $0.5g,g$
Explanation:
Ans. 0.5 g, g
T = 600 N (Block is on the verge of getting lifted)
T - 400 = 40a
600 - 400 = 40a $a = \frac{200}{40} = 5$ ; $\frac{g}{2} = 0.5g$ If boy move with 2g acceleration $T - 40\ g = 40 \times 2g \Rightarrow T = 120\ g$ For block 120 g - 60 g = 60 a $q = q$
a = g
T = 600 N (Block is on the verge of getting lifted)
T - 400 = 40a
600 - 400 = 40a $a = \frac{200}{40} = 5$ ; $\frac{g}{2} = 0.5g$ If boy move with 2g acceleration $T - 40\ g = 40 \times 2g \Rightarrow T = 120\ g$ For block 120 g - 60 g = 60 a $q = q$
a = g
Q20
Allen
Newtons Laws of Motion
Numerical
In the system shown, pulley and strings are ideal. The vertically upward pull F is being increased gradually, find magnitude of F and acceleration of the 5 kg block at the moment the 10 kg block leaves the floor.
Correct Answer: $200 \mathrm{~N}, 10 \mathrm{~m} / \mathrm{s}^{2}$
Explanation:
Ans. 200 N, 10 m/s $^{2}$ When 10 kg leave contact $\frac{F}{2}=100\Rightarrow F=200N$ 100 - 50 = 5a
a = 10 m/s $^{2}$
a = 10 m/s $^{2}$
Q21
Allen
Newtons Laws of Motion
Numerical
Force F is applied on upper pulley. If $F = 30t N$ where t is time in second. Find the time when $m_{1}$ loses contact with floor.
Correct Answer: 2 sec
Explanation:
Ans. 2 sec
If 4 kg is leaving ground,
20t = 40
If 4 kg is leaving ground,
20t = 40
Q22
Allen
Newtons Laws of Motion
Numerical
In the figure shown, all surfaces are smooth and block A and wedge B have mass 10 kg and 20 kg respectively. Find normal reaction between block A and B, spring force and normal reaction of ground on block B. $(g = 10 \, m/s^{2})$ .
Correct Answer: 80 N, 48 N, 264 N
Explanation:
Ans. 80 N, 48 N, 264 N
$\begin{array}{c} {N = 1 0 0 \cos 3 7 = 1 0 0 \times \frac {4}{5}} \\ {= 8 0 N} \end{array}$
$\begin{array}{r l} N _ {B} & = 2 0 0 + 8 0 \cos 3 7 \\ & = 2 0 0 + 8 0 \times \frac {4}{5} \\ & = 2 6 4 N \end{array}$
Spring force $kx = 16 \times 3 = 48 \, N$ .
$\begin{array}{c} {N = 1 0 0 \cos 3 7 = 1 0 0 \times \frac {4}{5}} \\ {= 8 0 N} \end{array}$
$\begin{array}{r l} N _ {B} & = 2 0 0 + 8 0 \cos 3 7 \\ & = 2 0 0 + 8 0 \times \frac {4}{5} \\ & = 2 6 4 N \end{array}$
Spring force $kx = 16 \times 3 = 48 \, N$ .
Q23
Allen
Newtons Laws of Motion
Numerical
Find the initial acceleration of blocks A and B if the
(A) Upper spring is cut
(B) Lower spring is cut
(A) Upper spring is cut
(B) Lower spring is cut
Correct Answer: (a) $3g \downarrow, 0$ (b) $2g \uparrow, g \downarrow$
Explanation:
Ans. (a) $3g \downarrow, 0$ (b) $2g \uparrow, g \downarrow$
Initially
(a)
$a _ {m} = \frac {3 m g}{m} = 3 g \downarrow$
(b)
$a _ {m} = \frac {2 m g}{m} = 2 g \uparrow$
$a _ {2 m} = \frac {2 m g}{2 m} = g \downarrow$
Initially
(a)
$a _ {m} = \frac {3 m g}{m} = 3 g \downarrow$
(b)
$a _ {m} = \frac {2 m g}{m} = 2 g \uparrow$
$a _ {2 m} = \frac {2 m g}{2 m} = g \downarrow$
Q24
Allen
Newtons Laws of Motion
Numerical
The system shown in the figure is initially in equilibrium. A is of mass 2m and B, C, D and E are of mass m. Certain actions are performed on the system. Every action has been taken individually when the system is intact. Find the direction and magnitude of acceleration of the blocks after each of the following actions has been taken:
(I) Spring 1 is cut
(II) Spring 2 is cut
(III) String between C and D is cut.
(IV) String between $B$ and $C$ is cut.
(I) Spring 1 is cut
(II) Spring 2 is cut
(III) String between C and D is cut.
(IV) String between $B$ and $C$ is cut.
Correct Answer: (i) $a_{A} = g\downarrow ,a_{B} = \frac{2g}{3}\uparrow ,a_{c} = \frac{2g}{3}\downarrow ,a_{D} = \frac{2g}{3}\downarrow ,a_{E} = 0$
(ii) $a_{A} = 0$ , $a_{B} = \frac{g}{3}\downarrow$ , $a_{C} = \frac{g}{3}\uparrow$ , $a_{D} = \frac{g}{3}\uparrow$ , $a_{E} = g\downarrow$ ,
(iii) $a_{A} = 0, a_{B} = g \downarrow, a_{C} = g \uparrow, a_{D} = 2g \downarrow, a_{E} = 0$
(iv) $a_{A} = 0, a_{B} = 3g \downarrow, a_{C} = \frac{3g}{2} \downarrow, a_{D} = \frac{3g}{2} \downarrow, a_{E} = 0$
Explanation:
Ans. (i) $a_{A} = g \downarrow, a_{B} = \frac{2g}{3} \uparrow, a_{c} = \frac{2g}{3} \downarrow, a_{D} = \frac{2g}{3} \downarrow, a_{E} = 0$
(ii) $a_{A} = 0$ , $a_{B} = \frac{g}{3}\downarrow$ , $a_{C} = \frac{g}{3}\uparrow$ , $a_{D} = \frac{g}{3}\uparrow$ , $a_{E} = g\downarrow$ ,
(iii) $a_{A} = 0, a_{B} = g \downarrow, a_{C} = g \uparrow, a_{D} = 2g \downarrow, a_{E} = 0$
(iv) $a_{A} = 0, a_{B} = 3g \downarrow, a_{C} = \frac{3g}{2} \downarrow, a_{D} = \frac{3g}{2} \downarrow, a_{E} = 0$
(i) Spring 1 is cut
$\vec {a} _ {A} = g \downarrow$
Acceleration of B, C, D are a.
$m g - T _ {1} = m a$
$T _ {1} - m g - T _ {2} = m a$
$T _ {2} - 2 m g = m a$
- 2 m g = 3 m a
$a = \frac {2 g}{3}$
$a _ {B} = \frac {2 g}{3} \uparrow a _ {C} = \frac {2 g}{3} \downarrow a _ {D} = \frac {2 g}{3} \downarrow$
$a_{E} = 0$
(ii) Spring (2) is cut $a_{A} = 0$
Let acceleration on B, C, D be a
$3 m g - T _ {1} = m a \quad a _ {B} = \frac {g}{3} \downarrow$
$a _ {c} = \frac {g}{3} \uparrow$
$T _ {1} - T _ {2} - m g = m a$
$a _ {D} = g / 3 \uparrow$
$T _ {2} - m g = m a \qquad \vec {a} _ {E} = g \downarrow$
$a = \frac {g}{3}$
(iii) String between $C$ and $D$ is cut
$a _ {A} = 0$
$a _ {E} = 0$
$a _ {D} = 2 g \downarrow$
Let acceleration of $B$ and $C$ be $a$
$3 m g - T _ {1} = m a$
$a _ {B} = g \downarrow$
$T _ {1} - m g = m a$
$a _ {C} = g \uparrow$
(iv) Between $B$ and $C$ is cut
$a _ {A} = 0$
$a _ {E} = 0$
$\mathrm{For} a _ {B}$
$3 m g = m a _ {B}$
$a _ {B} = 3 g \downarrow$
Let acceleration of $C$ and $D$ be $a$
2 m g - T = m a
m g + T = m a
$a = \frac {3 g}{2}$
$a _ {\mathrm{c}} = \frac {3 g}{2} \downarrow$
(ii) $a_{A} = 0$ , $a_{B} = \frac{g}{3}\downarrow$ , $a_{C} = \frac{g}{3}\uparrow$ , $a_{D} = \frac{g}{3}\uparrow$ , $a_{E} = g\downarrow$ ,
(iii) $a_{A} = 0, a_{B} = g \downarrow, a_{C} = g \uparrow, a_{D} = 2g \downarrow, a_{E} = 0$
(iv) $a_{A} = 0, a_{B} = 3g \downarrow, a_{C} = \frac{3g}{2} \downarrow, a_{D} = \frac{3g}{2} \downarrow, a_{E} = 0$
(i) Spring 1 is cut
$\vec {a} _ {A} = g \downarrow$
Acceleration of B, C, D are a.
$m g - T _ {1} = m a$
$T _ {1} - m g - T _ {2} = m a$
$T _ {2} - 2 m g = m a$
- 2 m g = 3 m a
$a = \frac {2 g}{3}$
$a _ {B} = \frac {2 g}{3} \uparrow a _ {C} = \frac {2 g}{3} \downarrow a _ {D} = \frac {2 g}{3} \downarrow$
$a_{E} = 0$
(ii) Spring (2) is cut $a_{A} = 0$
Let acceleration on B, C, D be a
$3 m g - T _ {1} = m a \quad a _ {B} = \frac {g}{3} \downarrow$
$a _ {c} = \frac {g}{3} \uparrow$
$T _ {1} - T _ {2} - m g = m a$
$a _ {D} = g / 3 \uparrow$
$T _ {2} - m g = m a \qquad \vec {a} _ {E} = g \downarrow$
$a = \frac {g}{3}$
(iii) String between $C$ and $D$ is cut
$a _ {A} = 0$
$a _ {E} = 0$
$a _ {D} = 2 g \downarrow$
Let acceleration of $B$ and $C$ be $a$
$3 m g - T _ {1} = m a$
$a _ {B} = g \downarrow$
$T _ {1} - m g = m a$
$a _ {C} = g \uparrow$
(iv) Between $B$ and $C$ is cut
$a _ {A} = 0$
$a _ {E} = 0$
$\mathrm{For} a _ {B}$
$3 m g = m a _ {B}$
$a _ {B} = 3 g \downarrow$
Let acceleration of $C$ and $D$ be $a$
2 m g - T = m a
m g + T = m a
$a = \frac {3 g}{2}$
$a _ {\mathrm{c}} = \frac {3 g}{2} \downarrow$
Q25
Allen
Newtons Laws of Motion
Numerical
Two trolley A and B are moving with accelerations a and 2a respectively in the same direction. To an observer in trolley A, the magnitude of pseudo force acting on a block of mass m on the trolley B is
Correct Answer: ma
Explanation:
Ans. ma
$a _ {D} = \frac {3 g}{2} \downarrow$
Pseudo force = $m_{object} \times a_{observer}$
$a _ {D} = \frac {3 g}{2} \downarrow$
Pseudo force = $m_{object} \times a_{observer}$