03. Forces in Mechanism
Laws of Motion

146061 A constant force \(F=m_{2} g / 2\) is applied on the block of mass \(m_{1}\) as shown in fig. The string and the pulley are light and the surface of the table is smooth. The acceleration of \(\mathrm{m}_{1}\) is

1 \(\frac{m_{2} g}{2\left(m_{1}+m_{2}\right)}\) towards right
2 \(\frac{m_{2} g}{2\left(m_{1}-m_{2}\right)}\) towards left
3 \(\frac{\mathrm{m}_{2} \mathrm{~g}}{2\left(\mathrm{~m}_{2}-\mathrm{m}_{1}\right)}\) towards right
4 \(\frac{m_{2} g}{2\left(m_{2}+m_{1}\right)}\) towards left
Laws of Motion

146062 The tension of a stretched string is increased by \(69 \%\). In order to keep its frequency of vibration constant, its length must be increased by :

1 \(30 \%\)
2 \(20 \%\)
3 \(69 \%\)
4 \(\sqrt{69} \%\)
Laws of Motion

146063 In the system shown in figure \(M_{1}>M_{2}\) and pulley and threads are ideal. System is held at rest by thread \(B C\). Just after thread \(B C\) is burnt.

1 Acceleration \(M_{1}\) and \(M_{2}\) will be upward
2 Magnitude of acceleration of both masses will be \(\frac{M_{1}-M_{2}}{M_{1}+M_{2}} g\)
3 Acceleration of \(M_{1}\) and \(M_{2}\) will be equal to zero
4 Acceleration of \(\mathrm{M}_{1}\) will be equal to zero, which that of \(m_{2}\) will be \(\frac{M_{1}-M_{2}}{M_{2}} g\) upward
Laws of Motion

146064 A block \(A\) of mass \(m_{1}\) rests on a horizontal table. A light string connected to it passes over a frictionless pulley at the edge of table and from its other end another block \(B\) of mass \(\mathbf{m}_{2}\) is suspended. The coefficient of kinetic friction between the block and the table is \(\mu_{\mathrm{k}}\). When the block \(A\) is sliding on the table, the tension in the string is

1 \(\frac{\left(m_{2}+\mu_{k} m_{1}\right) g}{\left(m_{1}+m_{2}\right)}\)
2 \(\frac{\left(m_{2}-\mu_{k} m_{1}\right) g}{\left(m_{1}+m_{2}\right)}\)
3 \(\frac{m_{1} m_{2}\left(1+\mu_{k}\right) g}{\left(m_{1}+m_{2}\right)}\)
4 \(\frac{m_{1} m_{2}\left(1-\mu_{k}\right) g}{\left(m_{1}+m_{2}\right)}\)
Laws of Motion

146061 A constant force \(F=m_{2} g / 2\) is applied on the block of mass \(m_{1}\) as shown in fig. The string and the pulley are light and the surface of the table is smooth. The acceleration of \(\mathrm{m}_{1}\) is

1 \(\frac{m_{2} g}{2\left(m_{1}+m_{2}\right)}\) towards right
2 \(\frac{m_{2} g}{2\left(m_{1}-m_{2}\right)}\) towards left
3 \(\frac{\mathrm{m}_{2} \mathrm{~g}}{2\left(\mathrm{~m}_{2}-\mathrm{m}_{1}\right)}\) towards right
4 \(\frac{m_{2} g}{2\left(m_{2}+m_{1}\right)}\) towards left
Laws of Motion

146062 The tension of a stretched string is increased by \(69 \%\). In order to keep its frequency of vibration constant, its length must be increased by :

1 \(30 \%\)
2 \(20 \%\)
3 \(69 \%\)
4 \(\sqrt{69} \%\)
Laws of Motion

146063 In the system shown in figure \(M_{1}>M_{2}\) and pulley and threads are ideal. System is held at rest by thread \(B C\). Just after thread \(B C\) is burnt.

1 Acceleration \(M_{1}\) and \(M_{2}\) will be upward
2 Magnitude of acceleration of both masses will be \(\frac{M_{1}-M_{2}}{M_{1}+M_{2}} g\)
3 Acceleration of \(M_{1}\) and \(M_{2}\) will be equal to zero
4 Acceleration of \(\mathrm{M}_{1}\) will be equal to zero, which that of \(m_{2}\) will be \(\frac{M_{1}-M_{2}}{M_{2}} g\) upward
Laws of Motion

146064 A block \(A\) of mass \(m_{1}\) rests on a horizontal table. A light string connected to it passes over a frictionless pulley at the edge of table and from its other end another block \(B\) of mass \(\mathbf{m}_{2}\) is suspended. The coefficient of kinetic friction between the block and the table is \(\mu_{\mathrm{k}}\). When the block \(A\) is sliding on the table, the tension in the string is

1 \(\frac{\left(m_{2}+\mu_{k} m_{1}\right) g}{\left(m_{1}+m_{2}\right)}\)
2 \(\frac{\left(m_{2}-\mu_{k} m_{1}\right) g}{\left(m_{1}+m_{2}\right)}\)
3 \(\frac{m_{1} m_{2}\left(1+\mu_{k}\right) g}{\left(m_{1}+m_{2}\right)}\)
4 \(\frac{m_{1} m_{2}\left(1-\mu_{k}\right) g}{\left(m_{1}+m_{2}\right)}\)
Laws of Motion

146061 A constant force \(F=m_{2} g / 2\) is applied on the block of mass \(m_{1}\) as shown in fig. The string and the pulley are light and the surface of the table is smooth. The acceleration of \(\mathrm{m}_{1}\) is

1 \(\frac{m_{2} g}{2\left(m_{1}+m_{2}\right)}\) towards right
2 \(\frac{m_{2} g}{2\left(m_{1}-m_{2}\right)}\) towards left
3 \(\frac{\mathrm{m}_{2} \mathrm{~g}}{2\left(\mathrm{~m}_{2}-\mathrm{m}_{1}\right)}\) towards right
4 \(\frac{m_{2} g}{2\left(m_{2}+m_{1}\right)}\) towards left
Laws of Motion

146062 The tension of a stretched string is increased by \(69 \%\). In order to keep its frequency of vibration constant, its length must be increased by :

1 \(30 \%\)
2 \(20 \%\)
3 \(69 \%\)
4 \(\sqrt{69} \%\)
Laws of Motion

146063 In the system shown in figure \(M_{1}>M_{2}\) and pulley and threads are ideal. System is held at rest by thread \(B C\). Just after thread \(B C\) is burnt.

1 Acceleration \(M_{1}\) and \(M_{2}\) will be upward
2 Magnitude of acceleration of both masses will be \(\frac{M_{1}-M_{2}}{M_{1}+M_{2}} g\)
3 Acceleration of \(M_{1}\) and \(M_{2}\) will be equal to zero
4 Acceleration of \(\mathrm{M}_{1}\) will be equal to zero, which that of \(m_{2}\) will be \(\frac{M_{1}-M_{2}}{M_{2}} g\) upward
Laws of Motion

146064 A block \(A\) of mass \(m_{1}\) rests on a horizontal table. A light string connected to it passes over a frictionless pulley at the edge of table and from its other end another block \(B\) of mass \(\mathbf{m}_{2}\) is suspended. The coefficient of kinetic friction between the block and the table is \(\mu_{\mathrm{k}}\). When the block \(A\) is sliding on the table, the tension in the string is

1 \(\frac{\left(m_{2}+\mu_{k} m_{1}\right) g}{\left(m_{1}+m_{2}\right)}\)
2 \(\frac{\left(m_{2}-\mu_{k} m_{1}\right) g}{\left(m_{1}+m_{2}\right)}\)
3 \(\frac{m_{1} m_{2}\left(1+\mu_{k}\right) g}{\left(m_{1}+m_{2}\right)}\)
4 \(\frac{m_{1} m_{2}\left(1-\mu_{k}\right) g}{\left(m_{1}+m_{2}\right)}\)
Laws of Motion

146061 A constant force \(F=m_{2} g / 2\) is applied on the block of mass \(m_{1}\) as shown in fig. The string and the pulley are light and the surface of the table is smooth. The acceleration of \(\mathrm{m}_{1}\) is

1 \(\frac{m_{2} g}{2\left(m_{1}+m_{2}\right)}\) towards right
2 \(\frac{m_{2} g}{2\left(m_{1}-m_{2}\right)}\) towards left
3 \(\frac{\mathrm{m}_{2} \mathrm{~g}}{2\left(\mathrm{~m}_{2}-\mathrm{m}_{1}\right)}\) towards right
4 \(\frac{m_{2} g}{2\left(m_{2}+m_{1}\right)}\) towards left
Laws of Motion

146062 The tension of a stretched string is increased by \(69 \%\). In order to keep its frequency of vibration constant, its length must be increased by :

1 \(30 \%\)
2 \(20 \%\)
3 \(69 \%\)
4 \(\sqrt{69} \%\)
Laws of Motion

146063 In the system shown in figure \(M_{1}>M_{2}\) and pulley and threads are ideal. System is held at rest by thread \(B C\). Just after thread \(B C\) is burnt.

1 Acceleration \(M_{1}\) and \(M_{2}\) will be upward
2 Magnitude of acceleration of both masses will be \(\frac{M_{1}-M_{2}}{M_{1}+M_{2}} g\)
3 Acceleration of \(M_{1}\) and \(M_{2}\) will be equal to zero
4 Acceleration of \(\mathrm{M}_{1}\) will be equal to zero, which that of \(m_{2}\) will be \(\frac{M_{1}-M_{2}}{M_{2}} g\) upward
Laws of Motion

146064 A block \(A\) of mass \(m_{1}\) rests on a horizontal table. A light string connected to it passes over a frictionless pulley at the edge of table and from its other end another block \(B\) of mass \(\mathbf{m}_{2}\) is suspended. The coefficient of kinetic friction between the block and the table is \(\mu_{\mathrm{k}}\). When the block \(A\) is sliding on the table, the tension in the string is

1 \(\frac{\left(m_{2}+\mu_{k} m_{1}\right) g}{\left(m_{1}+m_{2}\right)}\)
2 \(\frac{\left(m_{2}-\mu_{k} m_{1}\right) g}{\left(m_{1}+m_{2}\right)}\)
3 \(\frac{m_{1} m_{2}\left(1+\mu_{k}\right) g}{\left(m_{1}+m_{2}\right)}\)
4 \(\frac{m_{1} m_{2}\left(1-\mu_{k}\right) g}{\left(m_{1}+m_{2}\right)}\)