355501
Water falls from a height of 60 \(m\) at the rate of 15 \(kg/s\) to operate a turbine. The losses due to frictional force are \(10 \%\) of the input energy. How much power is generated by the turbine? \(\left( {g = 10\;m/{s^2}} \right)\)
1 8.1 \(kW\)
2 12.3 \(kW\)
3 7.0 \(kW\)
4 10.2 \(kW\)
Explanation:
Power generated by Turbine \(=P\) \(\begin{aligned}& P=0.9\left(\dfrac{\Delta m}{\Delta t}\right) g h \\& P=0.9 \times 15 \times 10 \times 60 \\& =8.1 \mathrm{~kW}\end{aligned}\)
NEET - 2021
PHXI06:WORK ENERGY AND POWER
355502
At any instant of time \({t}\), the displacement of any particle is given by \({2 t-1}\) (SI unit) under the influence of force of \(5\,N\). The value of instantaneous power is (in \({S I}\) unit):
1 10
2 5
3 7
4 6
Explanation:
Given that \({x=2 t-1}\) \({v=\dfrac{d x}{d t}=2 {~ms}^{-1}}\) \({P=F \cdot v=2 \times 5=10 {~W}}\)
NEET - 2024
PHXI06:WORK ENERGY AND POWER
355503
Power supplied to a particle of mass 2 \(kg\) varies with time as \(P=\dfrac{3 t^{2}}{2}\) watt. Here \(t\) is in second. If velocity of particle at \(t=0\) is \(v = 0\), the velocity of particle at time \(t = 2s\) will be:
1 \(2\,m/s\)
2 \(2\,\sqrt {2\,} m/s\)
3 \(1\,m/s\)
4 \(4\,m/s\)
Explanation:
From work-energy theorem, \(\Delta KE = {W_{net}}\,or\,{K_f} - {K_i} = \int P dt\) \(\frac{1}{2}m{v^2} - 0 = \int_0^2 {\left( {\frac{3}{2}{t^2}} \right)} \,dt\,\,{\mathop{\rm or}\nolimits} \,{v^2} = \left[ {\frac{{{r^3}}}{2}} \right]_0^2\) \( \Rightarrow v = 2\,m/s\)
PHXI06:WORK ENERGY AND POWER
355504
A car of mass ' \(m\) ' is driven with acceleration ' \(a\) ' along a straight level road against a constant external resistive force ' \(R\) '. When the velocity of the car is ' \(v\) ', the rate of which the engine of the car is doing work will be
1 \(m a v\)
2 \((m a-R) v\)
3 \(R v\)
4 \((R+m a) v\)
Explanation:
Force is required to oppose the resistive force \(\mathrm{R}\) and also to accelerate the body of mass \(\mathrm{m}\) with acceleration \(a\). Force \(F=R+m a\) \(\therefore\) Power \(=(\mathrm{R}+\mathrm{m} a) v\)
PHXI06:WORK ENERGY AND POWER
355505
A car of mass 500 \(kg\) is driven with acceleration \(1\;m/{s^2}\) along straight level road against constant external resistance of 1000 \(N\). When the velocity is 5 \(m/s\) the rate at which the engine is working is
1 7.5 \(kW\)
2 5 \(kW\)
3 10 \(kW\)
4 2.5 \(kW\)
Explanation:
Here the car is under acceleration \(P = Fv\quad \Rightarrow a = \frac{{F - {f_r}}}{m} \Rightarrow 1 = \frac{{F - 1000}}{{500}} \Rightarrow F = 1500 \Rightarrow P = 1500 \times 5 = 7500{\mkern 1mu} {\mkern 1mu} Watt\) \( \Rightarrow P = 1500 \times 5 = 7500\;Watt.\)
355501
Water falls from a height of 60 \(m\) at the rate of 15 \(kg/s\) to operate a turbine. The losses due to frictional force are \(10 \%\) of the input energy. How much power is generated by the turbine? \(\left( {g = 10\;m/{s^2}} \right)\)
1 8.1 \(kW\)
2 12.3 \(kW\)
3 7.0 \(kW\)
4 10.2 \(kW\)
Explanation:
Power generated by Turbine \(=P\) \(\begin{aligned}& P=0.9\left(\dfrac{\Delta m}{\Delta t}\right) g h \\& P=0.9 \times 15 \times 10 \times 60 \\& =8.1 \mathrm{~kW}\end{aligned}\)
NEET - 2021
PHXI06:WORK ENERGY AND POWER
355502
At any instant of time \({t}\), the displacement of any particle is given by \({2 t-1}\) (SI unit) under the influence of force of \(5\,N\). The value of instantaneous power is (in \({S I}\) unit):
1 10
2 5
3 7
4 6
Explanation:
Given that \({x=2 t-1}\) \({v=\dfrac{d x}{d t}=2 {~ms}^{-1}}\) \({P=F \cdot v=2 \times 5=10 {~W}}\)
NEET - 2024
PHXI06:WORK ENERGY AND POWER
355503
Power supplied to a particle of mass 2 \(kg\) varies with time as \(P=\dfrac{3 t^{2}}{2}\) watt. Here \(t\) is in second. If velocity of particle at \(t=0\) is \(v = 0\), the velocity of particle at time \(t = 2s\) will be:
1 \(2\,m/s\)
2 \(2\,\sqrt {2\,} m/s\)
3 \(1\,m/s\)
4 \(4\,m/s\)
Explanation:
From work-energy theorem, \(\Delta KE = {W_{net}}\,or\,{K_f} - {K_i} = \int P dt\) \(\frac{1}{2}m{v^2} - 0 = \int_0^2 {\left( {\frac{3}{2}{t^2}} \right)} \,dt\,\,{\mathop{\rm or}\nolimits} \,{v^2} = \left[ {\frac{{{r^3}}}{2}} \right]_0^2\) \( \Rightarrow v = 2\,m/s\)
PHXI06:WORK ENERGY AND POWER
355504
A car of mass ' \(m\) ' is driven with acceleration ' \(a\) ' along a straight level road against a constant external resistive force ' \(R\) '. When the velocity of the car is ' \(v\) ', the rate of which the engine of the car is doing work will be
1 \(m a v\)
2 \((m a-R) v\)
3 \(R v\)
4 \((R+m a) v\)
Explanation:
Force is required to oppose the resistive force \(\mathrm{R}\) and also to accelerate the body of mass \(\mathrm{m}\) with acceleration \(a\). Force \(F=R+m a\) \(\therefore\) Power \(=(\mathrm{R}+\mathrm{m} a) v\)
PHXI06:WORK ENERGY AND POWER
355505
A car of mass 500 \(kg\) is driven with acceleration \(1\;m/{s^2}\) along straight level road against constant external resistance of 1000 \(N\). When the velocity is 5 \(m/s\) the rate at which the engine is working is
1 7.5 \(kW\)
2 5 \(kW\)
3 10 \(kW\)
4 2.5 \(kW\)
Explanation:
Here the car is under acceleration \(P = Fv\quad \Rightarrow a = \frac{{F - {f_r}}}{m} \Rightarrow 1 = \frac{{F - 1000}}{{500}} \Rightarrow F = 1500 \Rightarrow P = 1500 \times 5 = 7500{\mkern 1mu} {\mkern 1mu} Watt\) \( \Rightarrow P = 1500 \times 5 = 7500\;Watt.\)
355501
Water falls from a height of 60 \(m\) at the rate of 15 \(kg/s\) to operate a turbine. The losses due to frictional force are \(10 \%\) of the input energy. How much power is generated by the turbine? \(\left( {g = 10\;m/{s^2}} \right)\)
1 8.1 \(kW\)
2 12.3 \(kW\)
3 7.0 \(kW\)
4 10.2 \(kW\)
Explanation:
Power generated by Turbine \(=P\) \(\begin{aligned}& P=0.9\left(\dfrac{\Delta m}{\Delta t}\right) g h \\& P=0.9 \times 15 \times 10 \times 60 \\& =8.1 \mathrm{~kW}\end{aligned}\)
NEET - 2021
PHXI06:WORK ENERGY AND POWER
355502
At any instant of time \({t}\), the displacement of any particle is given by \({2 t-1}\) (SI unit) under the influence of force of \(5\,N\). The value of instantaneous power is (in \({S I}\) unit):
1 10
2 5
3 7
4 6
Explanation:
Given that \({x=2 t-1}\) \({v=\dfrac{d x}{d t}=2 {~ms}^{-1}}\) \({P=F \cdot v=2 \times 5=10 {~W}}\)
NEET - 2024
PHXI06:WORK ENERGY AND POWER
355503
Power supplied to a particle of mass 2 \(kg\) varies with time as \(P=\dfrac{3 t^{2}}{2}\) watt. Here \(t\) is in second. If velocity of particle at \(t=0\) is \(v = 0\), the velocity of particle at time \(t = 2s\) will be:
1 \(2\,m/s\)
2 \(2\,\sqrt {2\,} m/s\)
3 \(1\,m/s\)
4 \(4\,m/s\)
Explanation:
From work-energy theorem, \(\Delta KE = {W_{net}}\,or\,{K_f} - {K_i} = \int P dt\) \(\frac{1}{2}m{v^2} - 0 = \int_0^2 {\left( {\frac{3}{2}{t^2}} \right)} \,dt\,\,{\mathop{\rm or}\nolimits} \,{v^2} = \left[ {\frac{{{r^3}}}{2}} \right]_0^2\) \( \Rightarrow v = 2\,m/s\)
PHXI06:WORK ENERGY AND POWER
355504
A car of mass ' \(m\) ' is driven with acceleration ' \(a\) ' along a straight level road against a constant external resistive force ' \(R\) '. When the velocity of the car is ' \(v\) ', the rate of which the engine of the car is doing work will be
1 \(m a v\)
2 \((m a-R) v\)
3 \(R v\)
4 \((R+m a) v\)
Explanation:
Force is required to oppose the resistive force \(\mathrm{R}\) and also to accelerate the body of mass \(\mathrm{m}\) with acceleration \(a\). Force \(F=R+m a\) \(\therefore\) Power \(=(\mathrm{R}+\mathrm{m} a) v\)
PHXI06:WORK ENERGY AND POWER
355505
A car of mass 500 \(kg\) is driven with acceleration \(1\;m/{s^2}\) along straight level road against constant external resistance of 1000 \(N\). When the velocity is 5 \(m/s\) the rate at which the engine is working is
1 7.5 \(kW\)
2 5 \(kW\)
3 10 \(kW\)
4 2.5 \(kW\)
Explanation:
Here the car is under acceleration \(P = Fv\quad \Rightarrow a = \frac{{F - {f_r}}}{m} \Rightarrow 1 = \frac{{F - 1000}}{{500}} \Rightarrow F = 1500 \Rightarrow P = 1500 \times 5 = 7500{\mkern 1mu} {\mkern 1mu} Watt\) \( \Rightarrow P = 1500 \times 5 = 7500\;Watt.\)
355501
Water falls from a height of 60 \(m\) at the rate of 15 \(kg/s\) to operate a turbine. The losses due to frictional force are \(10 \%\) of the input energy. How much power is generated by the turbine? \(\left( {g = 10\;m/{s^2}} \right)\)
1 8.1 \(kW\)
2 12.3 \(kW\)
3 7.0 \(kW\)
4 10.2 \(kW\)
Explanation:
Power generated by Turbine \(=P\) \(\begin{aligned}& P=0.9\left(\dfrac{\Delta m}{\Delta t}\right) g h \\& P=0.9 \times 15 \times 10 \times 60 \\& =8.1 \mathrm{~kW}\end{aligned}\)
NEET - 2021
PHXI06:WORK ENERGY AND POWER
355502
At any instant of time \({t}\), the displacement of any particle is given by \({2 t-1}\) (SI unit) under the influence of force of \(5\,N\). The value of instantaneous power is (in \({S I}\) unit):
1 10
2 5
3 7
4 6
Explanation:
Given that \({x=2 t-1}\) \({v=\dfrac{d x}{d t}=2 {~ms}^{-1}}\) \({P=F \cdot v=2 \times 5=10 {~W}}\)
NEET - 2024
PHXI06:WORK ENERGY AND POWER
355503
Power supplied to a particle of mass 2 \(kg\) varies with time as \(P=\dfrac{3 t^{2}}{2}\) watt. Here \(t\) is in second. If velocity of particle at \(t=0\) is \(v = 0\), the velocity of particle at time \(t = 2s\) will be:
1 \(2\,m/s\)
2 \(2\,\sqrt {2\,} m/s\)
3 \(1\,m/s\)
4 \(4\,m/s\)
Explanation:
From work-energy theorem, \(\Delta KE = {W_{net}}\,or\,{K_f} - {K_i} = \int P dt\) \(\frac{1}{2}m{v^2} - 0 = \int_0^2 {\left( {\frac{3}{2}{t^2}} \right)} \,dt\,\,{\mathop{\rm or}\nolimits} \,{v^2} = \left[ {\frac{{{r^3}}}{2}} \right]_0^2\) \( \Rightarrow v = 2\,m/s\)
PHXI06:WORK ENERGY AND POWER
355504
A car of mass ' \(m\) ' is driven with acceleration ' \(a\) ' along a straight level road against a constant external resistive force ' \(R\) '. When the velocity of the car is ' \(v\) ', the rate of which the engine of the car is doing work will be
1 \(m a v\)
2 \((m a-R) v\)
3 \(R v\)
4 \((R+m a) v\)
Explanation:
Force is required to oppose the resistive force \(\mathrm{R}\) and also to accelerate the body of mass \(\mathrm{m}\) with acceleration \(a\). Force \(F=R+m a\) \(\therefore\) Power \(=(\mathrm{R}+\mathrm{m} a) v\)
PHXI06:WORK ENERGY AND POWER
355505
A car of mass 500 \(kg\) is driven with acceleration \(1\;m/{s^2}\) along straight level road against constant external resistance of 1000 \(N\). When the velocity is 5 \(m/s\) the rate at which the engine is working is
1 7.5 \(kW\)
2 5 \(kW\)
3 10 \(kW\)
4 2.5 \(kW\)
Explanation:
Here the car is under acceleration \(P = Fv\quad \Rightarrow a = \frac{{F - {f_r}}}{m} \Rightarrow 1 = \frac{{F - 1000}}{{500}} \Rightarrow F = 1500 \Rightarrow P = 1500 \times 5 = 7500{\mkern 1mu} {\mkern 1mu} Watt\) \( \Rightarrow P = 1500 \times 5 = 7500\;Watt.\)
355501
Water falls from a height of 60 \(m\) at the rate of 15 \(kg/s\) to operate a turbine. The losses due to frictional force are \(10 \%\) of the input energy. How much power is generated by the turbine? \(\left( {g = 10\;m/{s^2}} \right)\)
1 8.1 \(kW\)
2 12.3 \(kW\)
3 7.0 \(kW\)
4 10.2 \(kW\)
Explanation:
Power generated by Turbine \(=P\) \(\begin{aligned}& P=0.9\left(\dfrac{\Delta m}{\Delta t}\right) g h \\& P=0.9 \times 15 \times 10 \times 60 \\& =8.1 \mathrm{~kW}\end{aligned}\)
NEET - 2021
PHXI06:WORK ENERGY AND POWER
355502
At any instant of time \({t}\), the displacement of any particle is given by \({2 t-1}\) (SI unit) under the influence of force of \(5\,N\). The value of instantaneous power is (in \({S I}\) unit):
1 10
2 5
3 7
4 6
Explanation:
Given that \({x=2 t-1}\) \({v=\dfrac{d x}{d t}=2 {~ms}^{-1}}\) \({P=F \cdot v=2 \times 5=10 {~W}}\)
NEET - 2024
PHXI06:WORK ENERGY AND POWER
355503
Power supplied to a particle of mass 2 \(kg\) varies with time as \(P=\dfrac{3 t^{2}}{2}\) watt. Here \(t\) is in second. If velocity of particle at \(t=0\) is \(v = 0\), the velocity of particle at time \(t = 2s\) will be:
1 \(2\,m/s\)
2 \(2\,\sqrt {2\,} m/s\)
3 \(1\,m/s\)
4 \(4\,m/s\)
Explanation:
From work-energy theorem, \(\Delta KE = {W_{net}}\,or\,{K_f} - {K_i} = \int P dt\) \(\frac{1}{2}m{v^2} - 0 = \int_0^2 {\left( {\frac{3}{2}{t^2}} \right)} \,dt\,\,{\mathop{\rm or}\nolimits} \,{v^2} = \left[ {\frac{{{r^3}}}{2}} \right]_0^2\) \( \Rightarrow v = 2\,m/s\)
PHXI06:WORK ENERGY AND POWER
355504
A car of mass ' \(m\) ' is driven with acceleration ' \(a\) ' along a straight level road against a constant external resistive force ' \(R\) '. When the velocity of the car is ' \(v\) ', the rate of which the engine of the car is doing work will be
1 \(m a v\)
2 \((m a-R) v\)
3 \(R v\)
4 \((R+m a) v\)
Explanation:
Force is required to oppose the resistive force \(\mathrm{R}\) and also to accelerate the body of mass \(\mathrm{m}\) with acceleration \(a\). Force \(F=R+m a\) \(\therefore\) Power \(=(\mathrm{R}+\mathrm{m} a) v\)
PHXI06:WORK ENERGY AND POWER
355505
A car of mass 500 \(kg\) is driven with acceleration \(1\;m/{s^2}\) along straight level road against constant external resistance of 1000 \(N\). When the velocity is 5 \(m/s\) the rate at which the engine is working is
1 7.5 \(kW\)
2 5 \(kW\)
3 10 \(kW\)
4 2.5 \(kW\)
Explanation:
Here the car is under acceleration \(P = Fv\quad \Rightarrow a = \frac{{F - {f_r}}}{m} \Rightarrow 1 = \frac{{F - 1000}}{{500}} \Rightarrow F = 1500 \Rightarrow P = 1500 \times 5 = 7500{\mkern 1mu} {\mkern 1mu} Watt\) \( \Rightarrow P = 1500 \times 5 = 7500\;Watt.\)
