160795
Three masses $m, 2 m$ and $3 m$ are attached with light string passing over a fixed frictionless pulley as shown in the figure. The tension in the string between $2 \mathrm{~m}$ and $3 \mathrm{~m}$ is ( $\mathrm{g}$ is acceleration due to gravity)
1 $2 \mathrm{mg}$
2 $3 \mathrm{mg}$
3 $6 \mathrm{mg}$
4 $1 \mathrm{mg}$
Explanation:
From figure, tension between masses $2 \mathrm{~m}$ and $3 \mathrm{~m}$ is $T_2$. $\because \mathrm{T}_2=\left(\frac{2 \mathrm{~m}_1 \mathrm{~m}_3}{\mathrm{~m}_1+\mathrm{m}_2+\mathrm{m}_3}\right) \mathrm{g}$ $\therefore \quad \mathrm{T}_2=\frac{2(\mathrm{~m})(3 \mathrm{~m})}{6 \mathrm{~m}} \mathrm{~g}=\mathrm{mg}$
NCERT-I-99
2 RBTS PAPER
160796
If $\mu_k$ is the coefficient of kinetic friction and $\mu_s$ is the coefficient of static friction then generally :
1 $\mu_k<\mu_s$
2 $\mu_s<\mu_k$
3 $\mu_{\mathrm{k}}=\mu_{\mathrm{s}}$
4 $\mu_s \leq \mu_k$
Explanation:
$\mu_{\mathrm{k}}<\mu_{\mathrm{s}}$ (coefficient of kinetic friction is generally less than coefficient of static friction)
NCERT-112
2 RBTS PAPER
160797
During padding of a bicycle, the force of friction exerted by the ground on the two wheels is such that it acts
1 In the forward direction on both the front and the rear wheels
2 In the backward direction on both the front and the rear wheels
3 In the forward direction on the front wheel and in the backward direction on the rear wheel
4 In the backward direction on the front wheel and in the forward direction on the rear wheel
Explanation:
In the backward direction on the front wheel and in the forward direction on the rear wheel
NCERT-I-101
2 RBTS PAPER
160798
Two blocks of masses $m$ and $2 m$ are placed on two smooth inclined planes joined back to back as shown in the figure. The masses will move with constant velocity if angle $\alpha$ is:
1 $\sin ^{-1}\left[\frac{1}{\sqrt{2}}\right]$
2 $\sin ^{-1}\left[\frac{1}{2 \sqrt{2}}\right]$
3 $\cos ^{-1}\left[\frac{\sqrt{3}}{4}\right]$
4 $\cos ^{-1}\left[\frac{1}{2 \sqrt{2}}\right]$
Explanation:
For constant velocity, acceleration of the blocks will be zero. $\Rightarrow \quad a=\frac{m_2 \sin \beta-m_1 \sin \alpha}{m_1+m_2} g=0$ $\therefore \quad m \sin \left(45^{\circ}\right)=2 m \sin \alpha$ $\frac{1}{2 \sqrt{2}}=\sin \alpha$ $\therefore \quad \alpha=\sin ^{-1}\left[\frac{1}{2 \sqrt{2}}\right]$
NCERT-I-99
2 RBTS PAPER
160799
Two masses of $4 \mathbf{k g}$ and $5 \mathbf{~ k g}$ are connected by a string passing through a frictionless pulley and are kept on a frictionless table as shown in the figure. The acceleration of $5 \mathbf{~ k g}$ mass is :
1 $49 \mathrm{~m} / \mathrm{s}^2$
2 $5.44 \mathrm{~m} / \mathrm{s}^2$
3 $19.5 \mathrm{~m} / \mathrm{s}^2$
4 $2.72 \mathrm{~m} / \mathrm{s}^2$
Explanation:
For $5 \mathrm{~kg}$ mass, $5 \mathrm{~g}-\mathrm{T}=5 \mathrm{a}$ and for $4 \mathrm{~kg}$ mass $T=4 a$ Adding (i) and (ii) $9 \mathrm{a}=5 \mathrm{~g}, \mathrm{a}=\frac{5 \times 9.8}{9}$ $=\frac{49}{9}=5.44 \mathrm{~ms}^{-2}$
160795
Three masses $m, 2 m$ and $3 m$ are attached with light string passing over a fixed frictionless pulley as shown in the figure. The tension in the string between $2 \mathrm{~m}$ and $3 \mathrm{~m}$ is ( $\mathrm{g}$ is acceleration due to gravity)
1 $2 \mathrm{mg}$
2 $3 \mathrm{mg}$
3 $6 \mathrm{mg}$
4 $1 \mathrm{mg}$
Explanation:
From figure, tension between masses $2 \mathrm{~m}$ and $3 \mathrm{~m}$ is $T_2$. $\because \mathrm{T}_2=\left(\frac{2 \mathrm{~m}_1 \mathrm{~m}_3}{\mathrm{~m}_1+\mathrm{m}_2+\mathrm{m}_3}\right) \mathrm{g}$ $\therefore \quad \mathrm{T}_2=\frac{2(\mathrm{~m})(3 \mathrm{~m})}{6 \mathrm{~m}} \mathrm{~g}=\mathrm{mg}$
NCERT-I-99
2 RBTS PAPER
160796
If $\mu_k$ is the coefficient of kinetic friction and $\mu_s$ is the coefficient of static friction then generally :
1 $\mu_k<\mu_s$
2 $\mu_s<\mu_k$
3 $\mu_{\mathrm{k}}=\mu_{\mathrm{s}}$
4 $\mu_s \leq \mu_k$
Explanation:
$\mu_{\mathrm{k}}<\mu_{\mathrm{s}}$ (coefficient of kinetic friction is generally less than coefficient of static friction)
NCERT-112
2 RBTS PAPER
160797
During padding of a bicycle, the force of friction exerted by the ground on the two wheels is such that it acts
1 In the forward direction on both the front and the rear wheels
2 In the backward direction on both the front and the rear wheels
3 In the forward direction on the front wheel and in the backward direction on the rear wheel
4 In the backward direction on the front wheel and in the forward direction on the rear wheel
Explanation:
In the backward direction on the front wheel and in the forward direction on the rear wheel
NCERT-I-101
2 RBTS PAPER
160798
Two blocks of masses $m$ and $2 m$ are placed on two smooth inclined planes joined back to back as shown in the figure. The masses will move with constant velocity if angle $\alpha$ is:
1 $\sin ^{-1}\left[\frac{1}{\sqrt{2}}\right]$
2 $\sin ^{-1}\left[\frac{1}{2 \sqrt{2}}\right]$
3 $\cos ^{-1}\left[\frac{\sqrt{3}}{4}\right]$
4 $\cos ^{-1}\left[\frac{1}{2 \sqrt{2}}\right]$
Explanation:
For constant velocity, acceleration of the blocks will be zero. $\Rightarrow \quad a=\frac{m_2 \sin \beta-m_1 \sin \alpha}{m_1+m_2} g=0$ $\therefore \quad m \sin \left(45^{\circ}\right)=2 m \sin \alpha$ $\frac{1}{2 \sqrt{2}}=\sin \alpha$ $\therefore \quad \alpha=\sin ^{-1}\left[\frac{1}{2 \sqrt{2}}\right]$
NCERT-I-99
2 RBTS PAPER
160799
Two masses of $4 \mathbf{k g}$ and $5 \mathbf{~ k g}$ are connected by a string passing through a frictionless pulley and are kept on a frictionless table as shown in the figure. The acceleration of $5 \mathbf{~ k g}$ mass is :
1 $49 \mathrm{~m} / \mathrm{s}^2$
2 $5.44 \mathrm{~m} / \mathrm{s}^2$
3 $19.5 \mathrm{~m} / \mathrm{s}^2$
4 $2.72 \mathrm{~m} / \mathrm{s}^2$
Explanation:
For $5 \mathrm{~kg}$ mass, $5 \mathrm{~g}-\mathrm{T}=5 \mathrm{a}$ and for $4 \mathrm{~kg}$ mass $T=4 a$ Adding (i) and (ii) $9 \mathrm{a}=5 \mathrm{~g}, \mathrm{a}=\frac{5 \times 9.8}{9}$ $=\frac{49}{9}=5.44 \mathrm{~ms}^{-2}$
160795
Three masses $m, 2 m$ and $3 m$ are attached with light string passing over a fixed frictionless pulley as shown in the figure. The tension in the string between $2 \mathrm{~m}$ and $3 \mathrm{~m}$ is ( $\mathrm{g}$ is acceleration due to gravity)
1 $2 \mathrm{mg}$
2 $3 \mathrm{mg}$
3 $6 \mathrm{mg}$
4 $1 \mathrm{mg}$
Explanation:
From figure, tension between masses $2 \mathrm{~m}$ and $3 \mathrm{~m}$ is $T_2$. $\because \mathrm{T}_2=\left(\frac{2 \mathrm{~m}_1 \mathrm{~m}_3}{\mathrm{~m}_1+\mathrm{m}_2+\mathrm{m}_3}\right) \mathrm{g}$ $\therefore \quad \mathrm{T}_2=\frac{2(\mathrm{~m})(3 \mathrm{~m})}{6 \mathrm{~m}} \mathrm{~g}=\mathrm{mg}$
NCERT-I-99
2 RBTS PAPER
160796
If $\mu_k$ is the coefficient of kinetic friction and $\mu_s$ is the coefficient of static friction then generally :
1 $\mu_k<\mu_s$
2 $\mu_s<\mu_k$
3 $\mu_{\mathrm{k}}=\mu_{\mathrm{s}}$
4 $\mu_s \leq \mu_k$
Explanation:
$\mu_{\mathrm{k}}<\mu_{\mathrm{s}}$ (coefficient of kinetic friction is generally less than coefficient of static friction)
NCERT-112
2 RBTS PAPER
160797
During padding of a bicycle, the force of friction exerted by the ground on the two wheels is such that it acts
1 In the forward direction on both the front and the rear wheels
2 In the backward direction on both the front and the rear wheels
3 In the forward direction on the front wheel and in the backward direction on the rear wheel
4 In the backward direction on the front wheel and in the forward direction on the rear wheel
Explanation:
In the backward direction on the front wheel and in the forward direction on the rear wheel
NCERT-I-101
2 RBTS PAPER
160798
Two blocks of masses $m$ and $2 m$ are placed on two smooth inclined planes joined back to back as shown in the figure. The masses will move with constant velocity if angle $\alpha$ is:
1 $\sin ^{-1}\left[\frac{1}{\sqrt{2}}\right]$
2 $\sin ^{-1}\left[\frac{1}{2 \sqrt{2}}\right]$
3 $\cos ^{-1}\left[\frac{\sqrt{3}}{4}\right]$
4 $\cos ^{-1}\left[\frac{1}{2 \sqrt{2}}\right]$
Explanation:
For constant velocity, acceleration of the blocks will be zero. $\Rightarrow \quad a=\frac{m_2 \sin \beta-m_1 \sin \alpha}{m_1+m_2} g=0$ $\therefore \quad m \sin \left(45^{\circ}\right)=2 m \sin \alpha$ $\frac{1}{2 \sqrt{2}}=\sin \alpha$ $\therefore \quad \alpha=\sin ^{-1}\left[\frac{1}{2 \sqrt{2}}\right]$
NCERT-I-99
2 RBTS PAPER
160799
Two masses of $4 \mathbf{k g}$ and $5 \mathbf{~ k g}$ are connected by a string passing through a frictionless pulley and are kept on a frictionless table as shown in the figure. The acceleration of $5 \mathbf{~ k g}$ mass is :
1 $49 \mathrm{~m} / \mathrm{s}^2$
2 $5.44 \mathrm{~m} / \mathrm{s}^2$
3 $19.5 \mathrm{~m} / \mathrm{s}^2$
4 $2.72 \mathrm{~m} / \mathrm{s}^2$
Explanation:
For $5 \mathrm{~kg}$ mass, $5 \mathrm{~g}-\mathrm{T}=5 \mathrm{a}$ and for $4 \mathrm{~kg}$ mass $T=4 a$ Adding (i) and (ii) $9 \mathrm{a}=5 \mathrm{~g}, \mathrm{a}=\frac{5 \times 9.8}{9}$ $=\frac{49}{9}=5.44 \mathrm{~ms}^{-2}$
160795
Three masses $m, 2 m$ and $3 m$ are attached with light string passing over a fixed frictionless pulley as shown in the figure. The tension in the string between $2 \mathrm{~m}$ and $3 \mathrm{~m}$ is ( $\mathrm{g}$ is acceleration due to gravity)
1 $2 \mathrm{mg}$
2 $3 \mathrm{mg}$
3 $6 \mathrm{mg}$
4 $1 \mathrm{mg}$
Explanation:
From figure, tension between masses $2 \mathrm{~m}$ and $3 \mathrm{~m}$ is $T_2$. $\because \mathrm{T}_2=\left(\frac{2 \mathrm{~m}_1 \mathrm{~m}_3}{\mathrm{~m}_1+\mathrm{m}_2+\mathrm{m}_3}\right) \mathrm{g}$ $\therefore \quad \mathrm{T}_2=\frac{2(\mathrm{~m})(3 \mathrm{~m})}{6 \mathrm{~m}} \mathrm{~g}=\mathrm{mg}$
NCERT-I-99
2 RBTS PAPER
160796
If $\mu_k$ is the coefficient of kinetic friction and $\mu_s$ is the coefficient of static friction then generally :
1 $\mu_k<\mu_s$
2 $\mu_s<\mu_k$
3 $\mu_{\mathrm{k}}=\mu_{\mathrm{s}}$
4 $\mu_s \leq \mu_k$
Explanation:
$\mu_{\mathrm{k}}<\mu_{\mathrm{s}}$ (coefficient of kinetic friction is generally less than coefficient of static friction)
NCERT-112
2 RBTS PAPER
160797
During padding of a bicycle, the force of friction exerted by the ground on the two wheels is such that it acts
1 In the forward direction on both the front and the rear wheels
2 In the backward direction on both the front and the rear wheels
3 In the forward direction on the front wheel and in the backward direction on the rear wheel
4 In the backward direction on the front wheel and in the forward direction on the rear wheel
Explanation:
In the backward direction on the front wheel and in the forward direction on the rear wheel
NCERT-I-101
2 RBTS PAPER
160798
Two blocks of masses $m$ and $2 m$ are placed on two smooth inclined planes joined back to back as shown in the figure. The masses will move with constant velocity if angle $\alpha$ is:
1 $\sin ^{-1}\left[\frac{1}{\sqrt{2}}\right]$
2 $\sin ^{-1}\left[\frac{1}{2 \sqrt{2}}\right]$
3 $\cos ^{-1}\left[\frac{\sqrt{3}}{4}\right]$
4 $\cos ^{-1}\left[\frac{1}{2 \sqrt{2}}\right]$
Explanation:
For constant velocity, acceleration of the blocks will be zero. $\Rightarrow \quad a=\frac{m_2 \sin \beta-m_1 \sin \alpha}{m_1+m_2} g=0$ $\therefore \quad m \sin \left(45^{\circ}\right)=2 m \sin \alpha$ $\frac{1}{2 \sqrt{2}}=\sin \alpha$ $\therefore \quad \alpha=\sin ^{-1}\left[\frac{1}{2 \sqrt{2}}\right]$
NCERT-I-99
2 RBTS PAPER
160799
Two masses of $4 \mathbf{k g}$ and $5 \mathbf{~ k g}$ are connected by a string passing through a frictionless pulley and are kept on a frictionless table as shown in the figure. The acceleration of $5 \mathbf{~ k g}$ mass is :
1 $49 \mathrm{~m} / \mathrm{s}^2$
2 $5.44 \mathrm{~m} / \mathrm{s}^2$
3 $19.5 \mathrm{~m} / \mathrm{s}^2$
4 $2.72 \mathrm{~m} / \mathrm{s}^2$
Explanation:
For $5 \mathrm{~kg}$ mass, $5 \mathrm{~g}-\mathrm{T}=5 \mathrm{a}$ and for $4 \mathrm{~kg}$ mass $T=4 a$ Adding (i) and (ii) $9 \mathrm{a}=5 \mathrm{~g}, \mathrm{a}=\frac{5 \times 9.8}{9}$ $=\frac{49}{9}=5.44 \mathrm{~ms}^{-2}$
160795
Three masses $m, 2 m$ and $3 m$ are attached with light string passing over a fixed frictionless pulley as shown in the figure. The tension in the string between $2 \mathrm{~m}$ and $3 \mathrm{~m}$ is ( $\mathrm{g}$ is acceleration due to gravity)
1 $2 \mathrm{mg}$
2 $3 \mathrm{mg}$
3 $6 \mathrm{mg}$
4 $1 \mathrm{mg}$
Explanation:
From figure, tension between masses $2 \mathrm{~m}$ and $3 \mathrm{~m}$ is $T_2$. $\because \mathrm{T}_2=\left(\frac{2 \mathrm{~m}_1 \mathrm{~m}_3}{\mathrm{~m}_1+\mathrm{m}_2+\mathrm{m}_3}\right) \mathrm{g}$ $\therefore \quad \mathrm{T}_2=\frac{2(\mathrm{~m})(3 \mathrm{~m})}{6 \mathrm{~m}} \mathrm{~g}=\mathrm{mg}$
NCERT-I-99
2 RBTS PAPER
160796
If $\mu_k$ is the coefficient of kinetic friction and $\mu_s$ is the coefficient of static friction then generally :
1 $\mu_k<\mu_s$
2 $\mu_s<\mu_k$
3 $\mu_{\mathrm{k}}=\mu_{\mathrm{s}}$
4 $\mu_s \leq \mu_k$
Explanation:
$\mu_{\mathrm{k}}<\mu_{\mathrm{s}}$ (coefficient of kinetic friction is generally less than coefficient of static friction)
NCERT-112
2 RBTS PAPER
160797
During padding of a bicycle, the force of friction exerted by the ground on the two wheels is such that it acts
1 In the forward direction on both the front and the rear wheels
2 In the backward direction on both the front and the rear wheels
3 In the forward direction on the front wheel and in the backward direction on the rear wheel
4 In the backward direction on the front wheel and in the forward direction on the rear wheel
Explanation:
In the backward direction on the front wheel and in the forward direction on the rear wheel
NCERT-I-101
2 RBTS PAPER
160798
Two blocks of masses $m$ and $2 m$ are placed on two smooth inclined planes joined back to back as shown in the figure. The masses will move with constant velocity if angle $\alpha$ is:
1 $\sin ^{-1}\left[\frac{1}{\sqrt{2}}\right]$
2 $\sin ^{-1}\left[\frac{1}{2 \sqrt{2}}\right]$
3 $\cos ^{-1}\left[\frac{\sqrt{3}}{4}\right]$
4 $\cos ^{-1}\left[\frac{1}{2 \sqrt{2}}\right]$
Explanation:
For constant velocity, acceleration of the blocks will be zero. $\Rightarrow \quad a=\frac{m_2 \sin \beta-m_1 \sin \alpha}{m_1+m_2} g=0$ $\therefore \quad m \sin \left(45^{\circ}\right)=2 m \sin \alpha$ $\frac{1}{2 \sqrt{2}}=\sin \alpha$ $\therefore \quad \alpha=\sin ^{-1}\left[\frac{1}{2 \sqrt{2}}\right]$
NCERT-I-99
2 RBTS PAPER
160799
Two masses of $4 \mathbf{k g}$ and $5 \mathbf{~ k g}$ are connected by a string passing through a frictionless pulley and are kept on a frictionless table as shown in the figure. The acceleration of $5 \mathbf{~ k g}$ mass is :
1 $49 \mathrm{~m} / \mathrm{s}^2$
2 $5.44 \mathrm{~m} / \mathrm{s}^2$
3 $19.5 \mathrm{~m} / \mathrm{s}^2$
4 $2.72 \mathrm{~m} / \mathrm{s}^2$
Explanation:
For $5 \mathrm{~kg}$ mass, $5 \mathrm{~g}-\mathrm{T}=5 \mathrm{a}$ and for $4 \mathrm{~kg}$ mass $T=4 a$ Adding (i) and (ii) $9 \mathrm{a}=5 \mathrm{~g}, \mathrm{a}=\frac{5 \times 9.8}{9}$ $=\frac{49}{9}=5.44 \mathrm{~ms}^{-2}$
