270309
The elevator shown in figure is descending with an acceleration of\(2 \mathrm{~m} / \mathrm{s}^{2}\). The mass of the block A \(=0.5 \mathrm{~kg}\). The force exerted by the block \(A\) on block \(B\) is
1 \(2 \mathrm{~N}\)
2 \(4 \mathrm{~N}\)
3 \(6 \mathrm{~N}\)
4 \(8 \mathrm{~N}\)
Explanation:
\(R=m(g-a)\)
Laws of Motion
270310
A block of mass \(m\) is pulled by a uniform chain of mass \(m\) tied to it by applying a force \(F\) at the other end of the chain. The tension at a point\(P\) which is at a distance of quarter of the length of the chain from the free end, will be
270311
Two masses of\(8 \mathrm{~kg}\) and \(4 \mathrm{~kg}\) are connected by a string as shown in figure over a frictionless pulley. The acceleration of the system is
1 \(4 m / s^{2}\)
2 \(2 \mathrm{~m} / \mathrm{s}^{2}\)
3 zero
4 \(9.8 \mathrm{~m} / \mathrm{s}^{2}\)
Explanation:
\(m_{2} g-T=m_{2} a ; T-m_{1} g \sin \theta=m_{1} a\)
Laws of Motion
270312
Consider the system shown in figure. The pulley and the string are light andall the surface are frictionless. The tension in the string is \(\left(\right.\) Take \(\left.g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\)
1 \(0 \mathrm{~N}\)
2 \(1 \mathrm{~N}\)
3 \(2 \mathrm{~N}\)
4 \(5 \mathrm{~N}\)
Explanation:
For the block on the surface\(T=m_{1} a\) For the hanging block \(m_{2} g-T=m_{2} a\)
270309
The elevator shown in figure is descending with an acceleration of\(2 \mathrm{~m} / \mathrm{s}^{2}\). The mass of the block A \(=0.5 \mathrm{~kg}\). The force exerted by the block \(A\) on block \(B\) is
1 \(2 \mathrm{~N}\)
2 \(4 \mathrm{~N}\)
3 \(6 \mathrm{~N}\)
4 \(8 \mathrm{~N}\)
Explanation:
\(R=m(g-a)\)
Laws of Motion
270310
A block of mass \(m\) is pulled by a uniform chain of mass \(m\) tied to it by applying a force \(F\) at the other end of the chain. The tension at a point\(P\) which is at a distance of quarter of the length of the chain from the free end, will be
270311
Two masses of\(8 \mathrm{~kg}\) and \(4 \mathrm{~kg}\) are connected by a string as shown in figure over a frictionless pulley. The acceleration of the system is
1 \(4 m / s^{2}\)
2 \(2 \mathrm{~m} / \mathrm{s}^{2}\)
3 zero
4 \(9.8 \mathrm{~m} / \mathrm{s}^{2}\)
Explanation:
\(m_{2} g-T=m_{2} a ; T-m_{1} g \sin \theta=m_{1} a\)
Laws of Motion
270312
Consider the system shown in figure. The pulley and the string are light andall the surface are frictionless. The tension in the string is \(\left(\right.\) Take \(\left.g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\)
1 \(0 \mathrm{~N}\)
2 \(1 \mathrm{~N}\)
3 \(2 \mathrm{~N}\)
4 \(5 \mathrm{~N}\)
Explanation:
For the block on the surface\(T=m_{1} a\) For the hanging block \(m_{2} g-T=m_{2} a\)
270309
The elevator shown in figure is descending with an acceleration of\(2 \mathrm{~m} / \mathrm{s}^{2}\). The mass of the block A \(=0.5 \mathrm{~kg}\). The force exerted by the block \(A\) on block \(B\) is
1 \(2 \mathrm{~N}\)
2 \(4 \mathrm{~N}\)
3 \(6 \mathrm{~N}\)
4 \(8 \mathrm{~N}\)
Explanation:
\(R=m(g-a)\)
Laws of Motion
270310
A block of mass \(m\) is pulled by a uniform chain of mass \(m\) tied to it by applying a force \(F\) at the other end of the chain. The tension at a point\(P\) which is at a distance of quarter of the length of the chain from the free end, will be
270311
Two masses of\(8 \mathrm{~kg}\) and \(4 \mathrm{~kg}\) are connected by a string as shown in figure over a frictionless pulley. The acceleration of the system is
1 \(4 m / s^{2}\)
2 \(2 \mathrm{~m} / \mathrm{s}^{2}\)
3 zero
4 \(9.8 \mathrm{~m} / \mathrm{s}^{2}\)
Explanation:
\(m_{2} g-T=m_{2} a ; T-m_{1} g \sin \theta=m_{1} a\)
Laws of Motion
270312
Consider the system shown in figure. The pulley and the string are light andall the surface are frictionless. The tension in the string is \(\left(\right.\) Take \(\left.g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\)
1 \(0 \mathrm{~N}\)
2 \(1 \mathrm{~N}\)
3 \(2 \mathrm{~N}\)
4 \(5 \mathrm{~N}\)
Explanation:
For the block on the surface\(T=m_{1} a\) For the hanging block \(m_{2} g-T=m_{2} a\)
270309
The elevator shown in figure is descending with an acceleration of\(2 \mathrm{~m} / \mathrm{s}^{2}\). The mass of the block A \(=0.5 \mathrm{~kg}\). The force exerted by the block \(A\) on block \(B\) is
1 \(2 \mathrm{~N}\)
2 \(4 \mathrm{~N}\)
3 \(6 \mathrm{~N}\)
4 \(8 \mathrm{~N}\)
Explanation:
\(R=m(g-a)\)
Laws of Motion
270310
A block of mass \(m\) is pulled by a uniform chain of mass \(m\) tied to it by applying a force \(F\) at the other end of the chain. The tension at a point\(P\) which is at a distance of quarter of the length of the chain from the free end, will be
270311
Two masses of\(8 \mathrm{~kg}\) and \(4 \mathrm{~kg}\) are connected by a string as shown in figure over a frictionless pulley. The acceleration of the system is
1 \(4 m / s^{2}\)
2 \(2 \mathrm{~m} / \mathrm{s}^{2}\)
3 zero
4 \(9.8 \mathrm{~m} / \mathrm{s}^{2}\)
Explanation:
\(m_{2} g-T=m_{2} a ; T-m_{1} g \sin \theta=m_{1} a\)
Laws of Motion
270312
Consider the system shown in figure. The pulley and the string are light andall the surface are frictionless. The tension in the string is \(\left(\right.\) Take \(\left.g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\)
1 \(0 \mathrm{~N}\)
2 \(1 \mathrm{~N}\)
3 \(2 \mathrm{~N}\)
4 \(5 \mathrm{~N}\)
Explanation:
For the block on the surface\(T=m_{1} a\) For the hanging block \(m_{2} g-T=m_{2} a\)