355484
If a machine gun fires \(n\) bullets per second each with kinetic energy \(K\), then the power of the machine gun is
1 \(\dfrac{K}{n}\)
2 \(n K^{2}\)
3 \(n K\)
4 \(n^{2} K\)
Explanation:
\(P = \frac{W}{t} \Rightarrow P = \frac{{nK}}{{1s}} = nK\)
PHXI06:WORK ENERGY AND POWER
355485
A motor pump lifts 6 tonns of water from a well of depth 25 \(m\) to the first floor of height 35 \(m\) from the ground floor in 20 minutes. The power of the pump (in \(k W\) ) is \([g = 10\,m{s^{ - 2}}]\)
355486
Power of a water pump is 2 \(kW\). If \(g = 10\;m/{\sec ^2}\), the amount of water it can raise in one minute to a height of 10 \(m\) is
1 100 litre
2 1200 litre
3 2000 litre
4 1000 litre
Explanation:
\(P = \frac{{mgh}}{t} \Rightarrow m = \frac{{p \times t}}{{gh}} = \frac{{2 \times {{10}^3} \times 60}}{{10 \times 10}} = 1200\;kg\) As volume \( = \frac{{ mass }}{{ density }} \Rightarrow V = \frac{{1200\;kg}}{{{{10}^3}\;kg/{m^3}}} = 1.2\;{m^3}\) Volume \( = 1.2\;{m^3} = 1.2 \times {10^3}{\mathop{\rm litre}\nolimits} {\rm{ }} = 1200\,\,{\mathop{\rm litre}\nolimits} .\)
PHXI06:WORK ENERGY AND POWER
355487
The ratio of powers of two motors is \(\dfrac{3 \sqrt{x}}{\sqrt{x}+1}\), that are capable of raising \(300\;kg\) water in 5 minutes and \(50\;kg\) water in 2 minutes respectively from a well of \(100\;m\) deep. The value of \(x\) will be
1 16
2 2
3 4
4 2.4
Explanation:
Power is given by \(P=\dfrac{W}{t}=\dfrac{m g h}{t}\) \(\frac{{{P_1}}}{{{P_2}}} = \frac{{{m_1}g{h_1}/{t_1}}}{{{m_2}g{h_2}/{t_2}}} = \frac{{{m_1}}}{{{m_2}}} \cdot \frac{{{t_2}}}{{{t_1}}}\) \( = \frac{{300}}{{50}} \times \frac{2}{5} = \frac{{12}}{5}\) \(\therefore \dfrac{P_{1}}{P_{2}}=\dfrac{3 \times 4}{4+1}=\dfrac{3 \times \sqrt{16}}{\sqrt{16}+1}\) \(=\dfrac{3 \times \sqrt{16}}{\sqrt{16}+1}\). On comparing with \(\dfrac{3 \sqrt{x}}{\sqrt{x}+1}\) \(\Rightarrow x=16\)
355484
If a machine gun fires \(n\) bullets per second each with kinetic energy \(K\), then the power of the machine gun is
1 \(\dfrac{K}{n}\)
2 \(n K^{2}\)
3 \(n K\)
4 \(n^{2} K\)
Explanation:
\(P = \frac{W}{t} \Rightarrow P = \frac{{nK}}{{1s}} = nK\)
PHXI06:WORK ENERGY AND POWER
355485
A motor pump lifts 6 tonns of water from a well of depth 25 \(m\) to the first floor of height 35 \(m\) from the ground floor in 20 minutes. The power of the pump (in \(k W\) ) is \([g = 10\,m{s^{ - 2}}]\)
355486
Power of a water pump is 2 \(kW\). If \(g = 10\;m/{\sec ^2}\), the amount of water it can raise in one minute to a height of 10 \(m\) is
1 100 litre
2 1200 litre
3 2000 litre
4 1000 litre
Explanation:
\(P = \frac{{mgh}}{t} \Rightarrow m = \frac{{p \times t}}{{gh}} = \frac{{2 \times {{10}^3} \times 60}}{{10 \times 10}} = 1200\;kg\) As volume \( = \frac{{ mass }}{{ density }} \Rightarrow V = \frac{{1200\;kg}}{{{{10}^3}\;kg/{m^3}}} = 1.2\;{m^3}\) Volume \( = 1.2\;{m^3} = 1.2 \times {10^3}{\mathop{\rm litre}\nolimits} {\rm{ }} = 1200\,\,{\mathop{\rm litre}\nolimits} .\)
PHXI06:WORK ENERGY AND POWER
355487
The ratio of powers of two motors is \(\dfrac{3 \sqrt{x}}{\sqrt{x}+1}\), that are capable of raising \(300\;kg\) water in 5 minutes and \(50\;kg\) water in 2 minutes respectively from a well of \(100\;m\) deep. The value of \(x\) will be
1 16
2 2
3 4
4 2.4
Explanation:
Power is given by \(P=\dfrac{W}{t}=\dfrac{m g h}{t}\) \(\frac{{{P_1}}}{{{P_2}}} = \frac{{{m_1}g{h_1}/{t_1}}}{{{m_2}g{h_2}/{t_2}}} = \frac{{{m_1}}}{{{m_2}}} \cdot \frac{{{t_2}}}{{{t_1}}}\) \( = \frac{{300}}{{50}} \times \frac{2}{5} = \frac{{12}}{5}\) \(\therefore \dfrac{P_{1}}{P_{2}}=\dfrac{3 \times 4}{4+1}=\dfrac{3 \times \sqrt{16}}{\sqrt{16}+1}\) \(=\dfrac{3 \times \sqrt{16}}{\sqrt{16}+1}\). On comparing with \(\dfrac{3 \sqrt{x}}{\sqrt{x}+1}\) \(\Rightarrow x=16\)
355484
If a machine gun fires \(n\) bullets per second each with kinetic energy \(K\), then the power of the machine gun is
1 \(\dfrac{K}{n}\)
2 \(n K^{2}\)
3 \(n K\)
4 \(n^{2} K\)
Explanation:
\(P = \frac{W}{t} \Rightarrow P = \frac{{nK}}{{1s}} = nK\)
PHXI06:WORK ENERGY AND POWER
355485
A motor pump lifts 6 tonns of water from a well of depth 25 \(m\) to the first floor of height 35 \(m\) from the ground floor in 20 minutes. The power of the pump (in \(k W\) ) is \([g = 10\,m{s^{ - 2}}]\)
355486
Power of a water pump is 2 \(kW\). If \(g = 10\;m/{\sec ^2}\), the amount of water it can raise in one minute to a height of 10 \(m\) is
1 100 litre
2 1200 litre
3 2000 litre
4 1000 litre
Explanation:
\(P = \frac{{mgh}}{t} \Rightarrow m = \frac{{p \times t}}{{gh}} = \frac{{2 \times {{10}^3} \times 60}}{{10 \times 10}} = 1200\;kg\) As volume \( = \frac{{ mass }}{{ density }} \Rightarrow V = \frac{{1200\;kg}}{{{{10}^3}\;kg/{m^3}}} = 1.2\;{m^3}\) Volume \( = 1.2\;{m^3} = 1.2 \times {10^3}{\mathop{\rm litre}\nolimits} {\rm{ }} = 1200\,\,{\mathop{\rm litre}\nolimits} .\)
PHXI06:WORK ENERGY AND POWER
355487
The ratio of powers of two motors is \(\dfrac{3 \sqrt{x}}{\sqrt{x}+1}\), that are capable of raising \(300\;kg\) water in 5 minutes and \(50\;kg\) water in 2 minutes respectively from a well of \(100\;m\) deep. The value of \(x\) will be
1 16
2 2
3 4
4 2.4
Explanation:
Power is given by \(P=\dfrac{W}{t}=\dfrac{m g h}{t}\) \(\frac{{{P_1}}}{{{P_2}}} = \frac{{{m_1}g{h_1}/{t_1}}}{{{m_2}g{h_2}/{t_2}}} = \frac{{{m_1}}}{{{m_2}}} \cdot \frac{{{t_2}}}{{{t_1}}}\) \( = \frac{{300}}{{50}} \times \frac{2}{5} = \frac{{12}}{5}\) \(\therefore \dfrac{P_{1}}{P_{2}}=\dfrac{3 \times 4}{4+1}=\dfrac{3 \times \sqrt{16}}{\sqrt{16}+1}\) \(=\dfrac{3 \times \sqrt{16}}{\sqrt{16}+1}\). On comparing with \(\dfrac{3 \sqrt{x}}{\sqrt{x}+1}\) \(\Rightarrow x=16\)
355484
If a machine gun fires \(n\) bullets per second each with kinetic energy \(K\), then the power of the machine gun is
1 \(\dfrac{K}{n}\)
2 \(n K^{2}\)
3 \(n K\)
4 \(n^{2} K\)
Explanation:
\(P = \frac{W}{t} \Rightarrow P = \frac{{nK}}{{1s}} = nK\)
PHXI06:WORK ENERGY AND POWER
355485
A motor pump lifts 6 tonns of water from a well of depth 25 \(m\) to the first floor of height 35 \(m\) from the ground floor in 20 minutes. The power of the pump (in \(k W\) ) is \([g = 10\,m{s^{ - 2}}]\)
355486
Power of a water pump is 2 \(kW\). If \(g = 10\;m/{\sec ^2}\), the amount of water it can raise in one minute to a height of 10 \(m\) is
1 100 litre
2 1200 litre
3 2000 litre
4 1000 litre
Explanation:
\(P = \frac{{mgh}}{t} \Rightarrow m = \frac{{p \times t}}{{gh}} = \frac{{2 \times {{10}^3} \times 60}}{{10 \times 10}} = 1200\;kg\) As volume \( = \frac{{ mass }}{{ density }} \Rightarrow V = \frac{{1200\;kg}}{{{{10}^3}\;kg/{m^3}}} = 1.2\;{m^3}\) Volume \( = 1.2\;{m^3} = 1.2 \times {10^3}{\mathop{\rm litre}\nolimits} {\rm{ }} = 1200\,\,{\mathop{\rm litre}\nolimits} .\)
PHXI06:WORK ENERGY AND POWER
355487
The ratio of powers of two motors is \(\dfrac{3 \sqrt{x}}{\sqrt{x}+1}\), that are capable of raising \(300\;kg\) water in 5 minutes and \(50\;kg\) water in 2 minutes respectively from a well of \(100\;m\) deep. The value of \(x\) will be
1 16
2 2
3 4
4 2.4
Explanation:
Power is given by \(P=\dfrac{W}{t}=\dfrac{m g h}{t}\) \(\frac{{{P_1}}}{{{P_2}}} = \frac{{{m_1}g{h_1}/{t_1}}}{{{m_2}g{h_2}/{t_2}}} = \frac{{{m_1}}}{{{m_2}}} \cdot \frac{{{t_2}}}{{{t_1}}}\) \( = \frac{{300}}{{50}} \times \frac{2}{5} = \frac{{12}}{5}\) \(\therefore \dfrac{P_{1}}{P_{2}}=\dfrac{3 \times 4}{4+1}=\dfrac{3 \times \sqrt{16}}{\sqrt{16}+1}\) \(=\dfrac{3 \times \sqrt{16}}{\sqrt{16}+1}\). On comparing with \(\dfrac{3 \sqrt{x}}{\sqrt{x}+1}\) \(\Rightarrow x=16\)
