160791
A $500 \mathrm{~kg}$ car takes a round turn of radius $50 \mathrm{~m}$ with a velocity of $36 \mathrm{~km} / \mathrm{hr}$. The centripetal force is
160792
A body takes time $t$ to reach the bottom of an inclined plane of angle $\theta$ with the horizontal. If the plane is made rough, time taken now is $2 t$. The coefficient of friction of the rough surface is
1 $\frac{2}{3} \tan \theta$
2 $\frac{1}{2} \tan \theta$
3 $\frac{3}{4} \tan \theta$
4 $\frac{1}{4} \tan \theta$
Explanation:
When there is no friction, $a=y \sin \theta$ When there is friction, $a^{\prime}=g(\sin \theta-\mu \cos \theta)$ If the length of the inclined plane is $d$, then $d=\frac{1}{2} a t^2=\frac{1}{2} a^{\prime}(2 t)^2$ or $\mathrm{a}=4 \mathrm{a}^{\prime}$ or $g \sin \theta=4 \mathrm{~g}(\sin \theta-\mu \cos \theta)$ $\sin \theta=4 \sin \theta-4 \mu \cos \theta$ $4 \mu \cos \theta=3 \sin \theta$ $\mu=\frac{3}{4} \tan \theta$
NCERT-I-101
2 RBTS PAPER
160793
A heavy uniform chain lies on horizontal table top. If the coefficient of friction between the chain and the table surface is $\mathbf{0 . 2 5}$, then the maximum fraction of the length of the chain that can hang over one edge of the table is
1 $20 \%$
2 $15 \%$
3 $25 \%$
4 $35 \%$
Explanation:
Suppose length $x$ of the chain hangs over one edge of the table. Mass per unit length of the chain $=\frac{M}{L}$ Weight of the hanging part, $W=\frac{M}{L} x g$ Weight of the hanging part applies force on the remaining part of the chain. The chain will not slide further if the limiting friction balances the weight W. $\therefore \quad \mathrm{f}_8^{\max }=\mathrm{W}=\frac{\mathrm{M}}{\mathrm{L}} \mathrm{xg}$ Normal reaction $=$ Weight of length $(L-x)$ of the chain $\mathrm{R}=\mathrm{W}^{\prime}=\frac{\mathrm{M}}{\mathrm{L}}(\mathrm{L}-\mathrm{x}) \mathrm{g}$ Now $\mathrm{f}_8^{\max }=\mu \mathrm{R}$ or $\frac{m}{L} x g=\mu \frac{M}{L}(L-x) g$ or $x=\mu(L-x)$ or $x=\frac{\mu L}{1+\mu}$ Given $\mu=0.25$ $\therefore \frac{\mathrm{x}}{\mathrm{L}} \times 100=\frac{0.25}{1+0.25} \times 100=20 \%$
NLI Expert
2 RBTS PAPER
160794
A man weighs $80 \mathrm{~kg}$. He stands on a weighing scale in a lift which is moving upwards with a uniform acceleration of $5 \mathrm{~m} / \mathrm{s}^2$. What would be the reading on the scale? $\left(g=10 \mathrm{~m} / \mathrm{s}^2\right)$
1 $400 \mathrm{~N}$
2 $800 \mathrm{~N}$
3 $1200 \mathrm{~N}$
4 zero
Explanation:
As the lift accelerates upward, the apparent weight, $R=m(g+a)=80(10+5)=1200 \mathrm{~N}$.
160791
A $500 \mathrm{~kg}$ car takes a round turn of radius $50 \mathrm{~m}$ with a velocity of $36 \mathrm{~km} / \mathrm{hr}$. The centripetal force is
160792
A body takes time $t$ to reach the bottom of an inclined plane of angle $\theta$ with the horizontal. If the plane is made rough, time taken now is $2 t$. The coefficient of friction of the rough surface is
1 $\frac{2}{3} \tan \theta$
2 $\frac{1}{2} \tan \theta$
3 $\frac{3}{4} \tan \theta$
4 $\frac{1}{4} \tan \theta$
Explanation:
When there is no friction, $a=y \sin \theta$ When there is friction, $a^{\prime}=g(\sin \theta-\mu \cos \theta)$ If the length of the inclined plane is $d$, then $d=\frac{1}{2} a t^2=\frac{1}{2} a^{\prime}(2 t)^2$ or $\mathrm{a}=4 \mathrm{a}^{\prime}$ or $g \sin \theta=4 \mathrm{~g}(\sin \theta-\mu \cos \theta)$ $\sin \theta=4 \sin \theta-4 \mu \cos \theta$ $4 \mu \cos \theta=3 \sin \theta$ $\mu=\frac{3}{4} \tan \theta$
NCERT-I-101
2 RBTS PAPER
160793
A heavy uniform chain lies on horizontal table top. If the coefficient of friction between the chain and the table surface is $\mathbf{0 . 2 5}$, then the maximum fraction of the length of the chain that can hang over one edge of the table is
1 $20 \%$
2 $15 \%$
3 $25 \%$
4 $35 \%$
Explanation:
Suppose length $x$ of the chain hangs over one edge of the table. Mass per unit length of the chain $=\frac{M}{L}$ Weight of the hanging part, $W=\frac{M}{L} x g$ Weight of the hanging part applies force on the remaining part of the chain. The chain will not slide further if the limiting friction balances the weight W. $\therefore \quad \mathrm{f}_8^{\max }=\mathrm{W}=\frac{\mathrm{M}}{\mathrm{L}} \mathrm{xg}$ Normal reaction $=$ Weight of length $(L-x)$ of the chain $\mathrm{R}=\mathrm{W}^{\prime}=\frac{\mathrm{M}}{\mathrm{L}}(\mathrm{L}-\mathrm{x}) \mathrm{g}$ Now $\mathrm{f}_8^{\max }=\mu \mathrm{R}$ or $\frac{m}{L} x g=\mu \frac{M}{L}(L-x) g$ or $x=\mu(L-x)$ or $x=\frac{\mu L}{1+\mu}$ Given $\mu=0.25$ $\therefore \frac{\mathrm{x}}{\mathrm{L}} \times 100=\frac{0.25}{1+0.25} \times 100=20 \%$
NLI Expert
2 RBTS PAPER
160794
A man weighs $80 \mathrm{~kg}$. He stands on a weighing scale in a lift which is moving upwards with a uniform acceleration of $5 \mathrm{~m} / \mathrm{s}^2$. What would be the reading on the scale? $\left(g=10 \mathrm{~m} / \mathrm{s}^2\right)$
1 $400 \mathrm{~N}$
2 $800 \mathrm{~N}$
3 $1200 \mathrm{~N}$
4 zero
Explanation:
As the lift accelerates upward, the apparent weight, $R=m(g+a)=80(10+5)=1200 \mathrm{~N}$.
160791
A $500 \mathrm{~kg}$ car takes a round turn of radius $50 \mathrm{~m}$ with a velocity of $36 \mathrm{~km} / \mathrm{hr}$. The centripetal force is
160792
A body takes time $t$ to reach the bottom of an inclined plane of angle $\theta$ with the horizontal. If the plane is made rough, time taken now is $2 t$. The coefficient of friction of the rough surface is
1 $\frac{2}{3} \tan \theta$
2 $\frac{1}{2} \tan \theta$
3 $\frac{3}{4} \tan \theta$
4 $\frac{1}{4} \tan \theta$
Explanation:
When there is no friction, $a=y \sin \theta$ When there is friction, $a^{\prime}=g(\sin \theta-\mu \cos \theta)$ If the length of the inclined plane is $d$, then $d=\frac{1}{2} a t^2=\frac{1}{2} a^{\prime}(2 t)^2$ or $\mathrm{a}=4 \mathrm{a}^{\prime}$ or $g \sin \theta=4 \mathrm{~g}(\sin \theta-\mu \cos \theta)$ $\sin \theta=4 \sin \theta-4 \mu \cos \theta$ $4 \mu \cos \theta=3 \sin \theta$ $\mu=\frac{3}{4} \tan \theta$
NCERT-I-101
2 RBTS PAPER
160793
A heavy uniform chain lies on horizontal table top. If the coefficient of friction between the chain and the table surface is $\mathbf{0 . 2 5}$, then the maximum fraction of the length of the chain that can hang over one edge of the table is
1 $20 \%$
2 $15 \%$
3 $25 \%$
4 $35 \%$
Explanation:
Suppose length $x$ of the chain hangs over one edge of the table. Mass per unit length of the chain $=\frac{M}{L}$ Weight of the hanging part, $W=\frac{M}{L} x g$ Weight of the hanging part applies force on the remaining part of the chain. The chain will not slide further if the limiting friction balances the weight W. $\therefore \quad \mathrm{f}_8^{\max }=\mathrm{W}=\frac{\mathrm{M}}{\mathrm{L}} \mathrm{xg}$ Normal reaction $=$ Weight of length $(L-x)$ of the chain $\mathrm{R}=\mathrm{W}^{\prime}=\frac{\mathrm{M}}{\mathrm{L}}(\mathrm{L}-\mathrm{x}) \mathrm{g}$ Now $\mathrm{f}_8^{\max }=\mu \mathrm{R}$ or $\frac{m}{L} x g=\mu \frac{M}{L}(L-x) g$ or $x=\mu(L-x)$ or $x=\frac{\mu L}{1+\mu}$ Given $\mu=0.25$ $\therefore \frac{\mathrm{x}}{\mathrm{L}} \times 100=\frac{0.25}{1+0.25} \times 100=20 \%$
NLI Expert
2 RBTS PAPER
160794
A man weighs $80 \mathrm{~kg}$. He stands on a weighing scale in a lift which is moving upwards with a uniform acceleration of $5 \mathrm{~m} / \mathrm{s}^2$. What would be the reading on the scale? $\left(g=10 \mathrm{~m} / \mathrm{s}^2\right)$
1 $400 \mathrm{~N}$
2 $800 \mathrm{~N}$
3 $1200 \mathrm{~N}$
4 zero
Explanation:
As the lift accelerates upward, the apparent weight, $R=m(g+a)=80(10+5)=1200 \mathrm{~N}$.
160791
A $500 \mathrm{~kg}$ car takes a round turn of radius $50 \mathrm{~m}$ with a velocity of $36 \mathrm{~km} / \mathrm{hr}$. The centripetal force is
160792
A body takes time $t$ to reach the bottom of an inclined plane of angle $\theta$ with the horizontal. If the plane is made rough, time taken now is $2 t$. The coefficient of friction of the rough surface is
1 $\frac{2}{3} \tan \theta$
2 $\frac{1}{2} \tan \theta$
3 $\frac{3}{4} \tan \theta$
4 $\frac{1}{4} \tan \theta$
Explanation:
When there is no friction, $a=y \sin \theta$ When there is friction, $a^{\prime}=g(\sin \theta-\mu \cos \theta)$ If the length of the inclined plane is $d$, then $d=\frac{1}{2} a t^2=\frac{1}{2} a^{\prime}(2 t)^2$ or $\mathrm{a}=4 \mathrm{a}^{\prime}$ or $g \sin \theta=4 \mathrm{~g}(\sin \theta-\mu \cos \theta)$ $\sin \theta=4 \sin \theta-4 \mu \cos \theta$ $4 \mu \cos \theta=3 \sin \theta$ $\mu=\frac{3}{4} \tan \theta$
NCERT-I-101
2 RBTS PAPER
160793
A heavy uniform chain lies on horizontal table top. If the coefficient of friction between the chain and the table surface is $\mathbf{0 . 2 5}$, then the maximum fraction of the length of the chain that can hang over one edge of the table is
1 $20 \%$
2 $15 \%$
3 $25 \%$
4 $35 \%$
Explanation:
Suppose length $x$ of the chain hangs over one edge of the table. Mass per unit length of the chain $=\frac{M}{L}$ Weight of the hanging part, $W=\frac{M}{L} x g$ Weight of the hanging part applies force on the remaining part of the chain. The chain will not slide further if the limiting friction balances the weight W. $\therefore \quad \mathrm{f}_8^{\max }=\mathrm{W}=\frac{\mathrm{M}}{\mathrm{L}} \mathrm{xg}$ Normal reaction $=$ Weight of length $(L-x)$ of the chain $\mathrm{R}=\mathrm{W}^{\prime}=\frac{\mathrm{M}}{\mathrm{L}}(\mathrm{L}-\mathrm{x}) \mathrm{g}$ Now $\mathrm{f}_8^{\max }=\mu \mathrm{R}$ or $\frac{m}{L} x g=\mu \frac{M}{L}(L-x) g$ or $x=\mu(L-x)$ or $x=\frac{\mu L}{1+\mu}$ Given $\mu=0.25$ $\therefore \frac{\mathrm{x}}{\mathrm{L}} \times 100=\frac{0.25}{1+0.25} \times 100=20 \%$
NLI Expert
2 RBTS PAPER
160794
A man weighs $80 \mathrm{~kg}$. He stands on a weighing scale in a lift which is moving upwards with a uniform acceleration of $5 \mathrm{~m} / \mathrm{s}^2$. What would be the reading on the scale? $\left(g=10 \mathrm{~m} / \mathrm{s}^2\right)$
1 $400 \mathrm{~N}$
2 $800 \mathrm{~N}$
3 $1200 \mathrm{~N}$
4 zero
Explanation:
As the lift accelerates upward, the apparent weight, $R=m(g+a)=80(10+5)=1200 \mathrm{~N}$.
