162571
A block having mass \(m\) collides with another stationary block having a mass \(\mathbf{2 m}\). The lighter block comes to rest after the collision. If the velocity of the first block is \(v\), then the value of coefficient of restitution will be
1 0.6
2 0.5
3 0.4
4 0.8
Explanation:
By conservation of momentum, \( m v=2 m v^{\prime} \Rightarrow v^{\prime}=\frac{v}{2} \)
Coefficient of restitution, \( e & =\frac{\text { Velocity of separation }}{\text { Velocity of approach }} \) \( =\frac{\mathrm{V}^{\prime}}{\mathrm{V}}=\frac{(\mathrm{v} / 2)}{\mathrm{v}}=\frac{1}{2}=0.5 \)
NCERT-I-84
3 RBTS PAPER
162572
A ball falls from rest from a height \(h\) on to a floor and rebounds to a height \(h / 4\). The coefficient of restitution between the ball and the floor is
162573
The block of mass \(M\) moving on the frictionless horizontal surface collides with the spring of spring constant \(\mathrm{K}\) and compresses it by length \(\mathrm{L}\). The maximum momentum of the block after collision is :
1 zero
2 \(\frac{M L^2}{K}\)
3 \(\sqrt{\mathrm{MK}} \mathrm{L}\)
4 \(\frac{K L^2}{2 M}\)
Explanation:
When block of mass \(\mathrm{M}\) collides with the spring its kinetic energy gets converted into elastic potential energy of the spring. From the law of conservation of energy \( \frac{1}{2} \mathrm{Mv}^2=\frac{1}{2} \mathrm{KL}^2 \quad \therefore \mathrm{v}=\sqrt{\frac{\mathrm{K}}{\mathrm{M}}} \mathrm{L} \)
Where \(v\) is the velocity of block by which it collides with spring. So its maximum momentum \( P=M v=M \sqrt{\frac{K}{M}} L=\sqrt{M K} L \)
After collision the block will rebound with same linear momentum.
NCERT-I-81
3 RBTS PAPER
162574
Water falls from a height of \(60 \mathrm{~m}\) at the rate of \(15 \mathrm{~kg} / \mathrm{s}\) to operate a turbine. The losses due to frictional forces are \(10 \%\) of energy. How much power is generated by the turbine: \(\left(g=10 \mathrm{~m} / \mathrm{s}^2\right)\)
1 \(12.3 \mathrm{~kW}\)
2 \(7.0 \mathrm{~kW}\)
3 \(8.1 \mathrm{~kW}\)
4 \(10.2 \mathrm{kw}\)
Explanation:
Power given to turbine \(=\frac{\mathrm{mgh}}{\mathrm{t}}\) \( P_{\text {in }}=\left(\frac{m}{t}\right) \times g \times h \Rightarrow P_{\text {in }}=15 \times 10 \times 60 \) \( \Rightarrow P_{\text {in }}=9000 \mathrm{~W} \Rightarrow P_{\text {in }}=9 \mathrm{~kW} \)
As efficiency of turbine is \(90 \%\) therefore power generated \(=90 \%\) of \(9 \mathrm{~kW}\) \( P_{\text {out }}=9 \times \frac{90}{100} \Rightarrow P_{\text {out }}=8.1 \mathrm{~kW} \)
162571
A block having mass \(m\) collides with another stationary block having a mass \(\mathbf{2 m}\). The lighter block comes to rest after the collision. If the velocity of the first block is \(v\), then the value of coefficient of restitution will be
1 0.6
2 0.5
3 0.4
4 0.8
Explanation:
By conservation of momentum, \( m v=2 m v^{\prime} \Rightarrow v^{\prime}=\frac{v}{2} \)
Coefficient of restitution, \( e & =\frac{\text { Velocity of separation }}{\text { Velocity of approach }} \) \( =\frac{\mathrm{V}^{\prime}}{\mathrm{V}}=\frac{(\mathrm{v} / 2)}{\mathrm{v}}=\frac{1}{2}=0.5 \)
NCERT-I-84
3 RBTS PAPER
162572
A ball falls from rest from a height \(h\) on to a floor and rebounds to a height \(h / 4\). The coefficient of restitution between the ball and the floor is
162573
The block of mass \(M\) moving on the frictionless horizontal surface collides with the spring of spring constant \(\mathrm{K}\) and compresses it by length \(\mathrm{L}\). The maximum momentum of the block after collision is :
1 zero
2 \(\frac{M L^2}{K}\)
3 \(\sqrt{\mathrm{MK}} \mathrm{L}\)
4 \(\frac{K L^2}{2 M}\)
Explanation:
When block of mass \(\mathrm{M}\) collides with the spring its kinetic energy gets converted into elastic potential energy of the spring. From the law of conservation of energy \( \frac{1}{2} \mathrm{Mv}^2=\frac{1}{2} \mathrm{KL}^2 \quad \therefore \mathrm{v}=\sqrt{\frac{\mathrm{K}}{\mathrm{M}}} \mathrm{L} \)
Where \(v\) is the velocity of block by which it collides with spring. So its maximum momentum \( P=M v=M \sqrt{\frac{K}{M}} L=\sqrt{M K} L \)
After collision the block will rebound with same linear momentum.
NCERT-I-81
3 RBTS PAPER
162574
Water falls from a height of \(60 \mathrm{~m}\) at the rate of \(15 \mathrm{~kg} / \mathrm{s}\) to operate a turbine. The losses due to frictional forces are \(10 \%\) of energy. How much power is generated by the turbine: \(\left(g=10 \mathrm{~m} / \mathrm{s}^2\right)\)
1 \(12.3 \mathrm{~kW}\)
2 \(7.0 \mathrm{~kW}\)
3 \(8.1 \mathrm{~kW}\)
4 \(10.2 \mathrm{kw}\)
Explanation:
Power given to turbine \(=\frac{\mathrm{mgh}}{\mathrm{t}}\) \( P_{\text {in }}=\left(\frac{m}{t}\right) \times g \times h \Rightarrow P_{\text {in }}=15 \times 10 \times 60 \) \( \Rightarrow P_{\text {in }}=9000 \mathrm{~W} \Rightarrow P_{\text {in }}=9 \mathrm{~kW} \)
As efficiency of turbine is \(90 \%\) therefore power generated \(=90 \%\) of \(9 \mathrm{~kW}\) \( P_{\text {out }}=9 \times \frac{90}{100} \Rightarrow P_{\text {out }}=8.1 \mathrm{~kW} \)
162571
A block having mass \(m\) collides with another stationary block having a mass \(\mathbf{2 m}\). The lighter block comes to rest after the collision. If the velocity of the first block is \(v\), then the value of coefficient of restitution will be
1 0.6
2 0.5
3 0.4
4 0.8
Explanation:
By conservation of momentum, \( m v=2 m v^{\prime} \Rightarrow v^{\prime}=\frac{v}{2} \)
Coefficient of restitution, \( e & =\frac{\text { Velocity of separation }}{\text { Velocity of approach }} \) \( =\frac{\mathrm{V}^{\prime}}{\mathrm{V}}=\frac{(\mathrm{v} / 2)}{\mathrm{v}}=\frac{1}{2}=0.5 \)
NCERT-I-84
3 RBTS PAPER
162572
A ball falls from rest from a height \(h\) on to a floor and rebounds to a height \(h / 4\). The coefficient of restitution between the ball and the floor is
162573
The block of mass \(M\) moving on the frictionless horizontal surface collides with the spring of spring constant \(\mathrm{K}\) and compresses it by length \(\mathrm{L}\). The maximum momentum of the block after collision is :
1 zero
2 \(\frac{M L^2}{K}\)
3 \(\sqrt{\mathrm{MK}} \mathrm{L}\)
4 \(\frac{K L^2}{2 M}\)
Explanation:
When block of mass \(\mathrm{M}\) collides with the spring its kinetic energy gets converted into elastic potential energy of the spring. From the law of conservation of energy \( \frac{1}{2} \mathrm{Mv}^2=\frac{1}{2} \mathrm{KL}^2 \quad \therefore \mathrm{v}=\sqrt{\frac{\mathrm{K}}{\mathrm{M}}} \mathrm{L} \)
Where \(v\) is the velocity of block by which it collides with spring. So its maximum momentum \( P=M v=M \sqrt{\frac{K}{M}} L=\sqrt{M K} L \)
After collision the block will rebound with same linear momentum.
NCERT-I-81
3 RBTS PAPER
162574
Water falls from a height of \(60 \mathrm{~m}\) at the rate of \(15 \mathrm{~kg} / \mathrm{s}\) to operate a turbine. The losses due to frictional forces are \(10 \%\) of energy. How much power is generated by the turbine: \(\left(g=10 \mathrm{~m} / \mathrm{s}^2\right)\)
1 \(12.3 \mathrm{~kW}\)
2 \(7.0 \mathrm{~kW}\)
3 \(8.1 \mathrm{~kW}\)
4 \(10.2 \mathrm{kw}\)
Explanation:
Power given to turbine \(=\frac{\mathrm{mgh}}{\mathrm{t}}\) \( P_{\text {in }}=\left(\frac{m}{t}\right) \times g \times h \Rightarrow P_{\text {in }}=15 \times 10 \times 60 \) \( \Rightarrow P_{\text {in }}=9000 \mathrm{~W} \Rightarrow P_{\text {in }}=9 \mathrm{~kW} \)
As efficiency of turbine is \(90 \%\) therefore power generated \(=90 \%\) of \(9 \mathrm{~kW}\) \( P_{\text {out }}=9 \times \frac{90}{100} \Rightarrow P_{\text {out }}=8.1 \mathrm{~kW} \)
162571
A block having mass \(m\) collides with another stationary block having a mass \(\mathbf{2 m}\). The lighter block comes to rest after the collision. If the velocity of the first block is \(v\), then the value of coefficient of restitution will be
1 0.6
2 0.5
3 0.4
4 0.8
Explanation:
By conservation of momentum, \( m v=2 m v^{\prime} \Rightarrow v^{\prime}=\frac{v}{2} \)
Coefficient of restitution, \( e & =\frac{\text { Velocity of separation }}{\text { Velocity of approach }} \) \( =\frac{\mathrm{V}^{\prime}}{\mathrm{V}}=\frac{(\mathrm{v} / 2)}{\mathrm{v}}=\frac{1}{2}=0.5 \)
NCERT-I-84
3 RBTS PAPER
162572
A ball falls from rest from a height \(h\) on to a floor and rebounds to a height \(h / 4\). The coefficient of restitution between the ball and the floor is
162573
The block of mass \(M\) moving on the frictionless horizontal surface collides with the spring of spring constant \(\mathrm{K}\) and compresses it by length \(\mathrm{L}\). The maximum momentum of the block after collision is :
1 zero
2 \(\frac{M L^2}{K}\)
3 \(\sqrt{\mathrm{MK}} \mathrm{L}\)
4 \(\frac{K L^2}{2 M}\)
Explanation:
When block of mass \(\mathrm{M}\) collides with the spring its kinetic energy gets converted into elastic potential energy of the spring. From the law of conservation of energy \( \frac{1}{2} \mathrm{Mv}^2=\frac{1}{2} \mathrm{KL}^2 \quad \therefore \mathrm{v}=\sqrt{\frac{\mathrm{K}}{\mathrm{M}}} \mathrm{L} \)
Where \(v\) is the velocity of block by which it collides with spring. So its maximum momentum \( P=M v=M \sqrt{\frac{K}{M}} L=\sqrt{M K} L \)
After collision the block will rebound with same linear momentum.
NCERT-I-81
3 RBTS PAPER
162574
Water falls from a height of \(60 \mathrm{~m}\) at the rate of \(15 \mathrm{~kg} / \mathrm{s}\) to operate a turbine. The losses due to frictional forces are \(10 \%\) of energy. How much power is generated by the turbine: \(\left(g=10 \mathrm{~m} / \mathrm{s}^2\right)\)
1 \(12.3 \mathrm{~kW}\)
2 \(7.0 \mathrm{~kW}\)
3 \(8.1 \mathrm{~kW}\)
4 \(10.2 \mathrm{kw}\)
Explanation:
Power given to turbine \(=\frac{\mathrm{mgh}}{\mathrm{t}}\) \( P_{\text {in }}=\left(\frac{m}{t}\right) \times g \times h \Rightarrow P_{\text {in }}=15 \times 10 \times 60 \) \( \Rightarrow P_{\text {in }}=9000 \mathrm{~W} \Rightarrow P_{\text {in }}=9 \mathrm{~kW} \)
As efficiency of turbine is \(90 \%\) therefore power generated \(=90 \%\) of \(9 \mathrm{~kW}\) \( P_{\text {out }}=9 \times \frac{90}{100} \Rightarrow P_{\text {out }}=8.1 \mathrm{~kW} \)
