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PHXI06:WORK ENERGY AND POWER

355488 Assertion :
Power of machine gun is determined by both, the number of bullets fired per second and kinetic energy of bullets.
Reason :
Power of any machine is defined as work done (by it) per unit time.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Kinetic energy of one bullet $= E_{k}$
Kinetic energy of $n$ bullet $= n E_{k}$
According to law of conservation of energy, the kinetic energy of bullets be equal to the work done by machine gun per sec.
Power $= \frac{work}{time} = n E_{k}$
So correct option is (1)

PHXI06:WORK ENERGY AND POWER

355489 A body is moved along a straight line by a machine delivering constant power. The distance moved by the body in time $t$ is proportional to

1

t^{2}

2

t^{3 / 2}

3

t^{1 / 2}

4

t^{3 / 4}

Explanation:

$P = F v = m a v = m (\frac{d v}{d t}) v \Rightarrow \frac{P}{m} d t = v d v$ $\Rightarrow \frac{P}{m} \times t = \frac{v^{2}}{2} \Rightarrow v = {(\frac{2 P}{m})}^{1 / 2} (t)^{1 / 2}$
Now $s = \int v d t = {(\frac{2 P}{m})}^{1 / 2} (t)^{1 / 2} d t$
$∴ s = {(\frac{2 P}{m})}^{1 / 2} [\frac{2 t^{3 / 2}}{3}] \Rightarrow s α t^{3 / 2}$

PHXI06:WORK ENERGY AND POWER

355490 A motor of power $p_{0}$ is used to deliver water at certain rate through a given horizontal pipe. To increase the rate of water through the same pipe $n$ times, the power of the motor is increased to $p_{1}$ . The ratio of $p_{1}$ to $p_{0}$ is

1

n^{4} : 1

2

n^{2} : 1

3

n : 1

4

n^{3} : 1

Explanation:

Power of motor initially $= p_{0}$
Let, rate of flow of motor $= (x)$
Since, power, $p_{0} = \frac{work}{time} = \frac{m g y}{t} = m g (\frac{y}{t})$
$\frac{y}{t} = x =$ rate of flow of water
$= m g x (1)$
If rate of flow of water is increased by $n$ times,i.e., $(n x)$
Increased power, $p_{1} = \frac{m g y^{'}}{t} = m g (\frac{y^{'}}{t})$ ,
$= nmg x$
The ratio of power
$\frac{p_{1}}{p_{0}} = \frac{n m g x}{m g x} = \frac{n}{1} \Rightarrow p_{1} : p_{0} \Rightarrow n : 1$

PHXI06:WORK ENERGY AND POWER

355491 A particle of mass $m$ is driven by a machine that delivers a constant power $K$ watts. If the particle starts from the rest the velocity of the particle at time $^{'} t^{'}$ is

1

\sqrt{\frac{2 K t}{m}}

2

\sqrt{2 K t}

3

\sqrt{\frac{m}{2 K t}}

4

\sqrt{2 K t m}

Explanation:

$p = F v = K$
$\begin{aligned} m \frac{d v}{d t} v = K \\ v d v = \frac{K}{m} d t \\ \frac{v^{2}}{2} = \frac{K}{m} t \Rightarrow v = \sqrt{\frac{2 K}{m}} t \end{aligned}$

PHXI06:WORK ENERGY AND POWER

355492 A motorcycle of mass $m$ resting on a frictionless road moves under the influence of a constant force $F$ . The work done by this force in moving the motorcycle is given by $F^{2} t^{2} / 2 m$ , where $t$ is the time in seconds.
Ratio of instantaneous power to average power of the motorcycle in $t = T$ second is:
supporting img

1

1 : 1

2

2 : 1

3

3 : 2

4

1 : 2

Explanation:

For instantaneous power
$\begin{aligned} P_{i n s} = \frac{d W}{d t} = \frac{F^{2}}{2 m} \cdot 2 t \Rightarrow P_{ins} = \frac{2 F^{2} T}{2 m} \\ P_{a v g} = \frac{Δ W}{T} = \frac{F^{2} (T^{2} - 0)}{2 m T} = \frac{F^{2} T}{2 m} \\ \frac{P_{ins}}{P_{avg}} = 2 \end{aligned}$

PHXI06:WORK ENERGY AND POWER

355488 Assertion :
Power of machine gun is determined by both, the number of bullets fired per second and kinetic energy of bullets.
Reason :
Power of any machine is defined as work done (by it) per unit time.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Kinetic energy of one bullet $= E_{k}$
Kinetic energy of $n$ bullet $= n E_{k}$
According to law of conservation of energy, the kinetic energy of bullets be equal to the work done by machine gun per sec.
Power $= \frac{work}{time} = n E_{k}$
So correct option is (1)

PHXI06:WORK ENERGY AND POWER

355489 A body is moved along a straight line by a machine delivering constant power. The distance moved by the body in time $t$ is proportional to

1

t^{2}

2

t^{3 / 2}

3

t^{1 / 2}

4

t^{3 / 4}

Explanation:

$P = F v = m a v = m (\frac{d v}{d t}) v \Rightarrow \frac{P}{m} d t = v d v$ $\Rightarrow \frac{P}{m} \times t = \frac{v^{2}}{2} \Rightarrow v = {(\frac{2 P}{m})}^{1 / 2} (t)^{1 / 2}$
Now $s = \int v d t = {(\frac{2 P}{m})}^{1 / 2} (t)^{1 / 2} d t$
$∴ s = {(\frac{2 P}{m})}^{1 / 2} [\frac{2 t^{3 / 2}}{3}] \Rightarrow s α t^{3 / 2}$

PHXI06:WORK ENERGY AND POWER

355490 A motor of power $p_{0}$ is used to deliver water at certain rate through a given horizontal pipe. To increase the rate of water through the same pipe $n$ times, the power of the motor is increased to $p_{1}$ . The ratio of $p_{1}$ to $p_{0}$ is

1

n^{4} : 1

2

n^{2} : 1

3

n : 1

4

n^{3} : 1

Explanation:

Power of motor initially $= p_{0}$
Let, rate of flow of motor $= (x)$
Since, power, $p_{0} = \frac{work}{time} = \frac{m g y}{t} = m g (\frac{y}{t})$
$\frac{y}{t} = x =$ rate of flow of water
$= m g x (1)$
If rate of flow of water is increased by $n$ times,i.e., $(n x)$
Increased power, $p_{1} = \frac{m g y^{'}}{t} = m g (\frac{y^{'}}{t})$ ,
$= nmg x$
The ratio of power
$\frac{p_{1}}{p_{0}} = \frac{n m g x}{m g x} = \frac{n}{1} \Rightarrow p_{1} : p_{0} \Rightarrow n : 1$

PHXI06:WORK ENERGY AND POWER

355491 A particle of mass $m$ is driven by a machine that delivers a constant power $K$ watts. If the particle starts from the rest the velocity of the particle at time $^{'} t^{'}$ is

1

\sqrt{\frac{2 K t}{m}}

2

\sqrt{2 K t}

3

\sqrt{\frac{m}{2 K t}}

4

\sqrt{2 K t m}

Explanation:

$p = F v = K$
$\begin{aligned} m \frac{d v}{d t} v = K \\ v d v = \frac{K}{m} d t \\ \frac{v^{2}}{2} = \frac{K}{m} t \Rightarrow v = \sqrt{\frac{2 K}{m}} t \end{aligned}$

PHXI06:WORK ENERGY AND POWER

355492 A motorcycle of mass $m$ resting on a frictionless road moves under the influence of a constant force $F$ . The work done by this force in moving the motorcycle is given by $F^{2} t^{2} / 2 m$ , where $t$ is the time in seconds.
Ratio of instantaneous power to average power of the motorcycle in $t = T$ second is:
supporting img

1

1 : 1

2

2 : 1

3

3 : 2

4

1 : 2

Explanation:

For instantaneous power
$\begin{aligned} P_{i n s} = \frac{d W}{d t} = \frac{F^{2}}{2 m} \cdot 2 t \Rightarrow P_{ins} = \frac{2 F^{2} T}{2 m} \\ P_{a v g} = \frac{Δ W}{T} = \frac{F^{2} (T^{2} - 0)}{2 m T} = \frac{F^{2} T}{2 m} \\ \frac{P_{ins}}{P_{avg}} = 2 \end{aligned}$

PHXI06:WORK ENERGY AND POWER

355488 Assertion :
Power of machine gun is determined by both, the number of bullets fired per second and kinetic energy of bullets.
Reason :
Power of any machine is defined as work done (by it) per unit time.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Kinetic energy of one bullet $= E_{k}$
Kinetic energy of $n$ bullet $= n E_{k}$
According to law of conservation of energy, the kinetic energy of bullets be equal to the work done by machine gun per sec.
Power $= \frac{work}{time} = n E_{k}$
So correct option is (1)

PHXI06:WORK ENERGY AND POWER

355489 A body is moved along a straight line by a machine delivering constant power. The distance moved by the body in time $t$ is proportional to

1

t^{2}

2

t^{3 / 2}

3

t^{1 / 2}

4

t^{3 / 4}

Explanation:

$P = F v = m a v = m (\frac{d v}{d t}) v \Rightarrow \frac{P}{m} d t = v d v$ $\Rightarrow \frac{P}{m} \times t = \frac{v^{2}}{2} \Rightarrow v = {(\frac{2 P}{m})}^{1 / 2} (t)^{1 / 2}$
Now $s = \int v d t = {(\frac{2 P}{m})}^{1 / 2} (t)^{1 / 2} d t$
$∴ s = {(\frac{2 P}{m})}^{1 / 2} [\frac{2 t^{3 / 2}}{3}] \Rightarrow s α t^{3 / 2}$

PHXI06:WORK ENERGY AND POWER

355490 A motor of power $p_{0}$ is used to deliver water at certain rate through a given horizontal pipe. To increase the rate of water through the same pipe $n$ times, the power of the motor is increased to $p_{1}$ . The ratio of $p_{1}$ to $p_{0}$ is

1

n^{4} : 1

2

n^{2} : 1

3

n : 1

4

n^{3} : 1

Explanation:

Power of motor initially $= p_{0}$
Let, rate of flow of motor $= (x)$
Since, power, $p_{0} = \frac{work}{time} = \frac{m g y}{t} = m g (\frac{y}{t})$
$\frac{y}{t} = x =$ rate of flow of water
$= m g x (1)$
If rate of flow of water is increased by $n$ times,i.e., $(n x)$
Increased power, $p_{1} = \frac{m g y^{'}}{t} = m g (\frac{y^{'}}{t})$ ,
$= nmg x$
The ratio of power
$\frac{p_{1}}{p_{0}} = \frac{n m g x}{m g x} = \frac{n}{1} \Rightarrow p_{1} : p_{0} \Rightarrow n : 1$

PHXI06:WORK ENERGY AND POWER

355491 A particle of mass $m$ is driven by a machine that delivers a constant power $K$ watts. If the particle starts from the rest the velocity of the particle at time $^{'} t^{'}$ is

1

\sqrt{\frac{2 K t}{m}}

2

\sqrt{2 K t}

3

\sqrt{\frac{m}{2 K t}}

4

\sqrt{2 K t m}

Explanation:

$p = F v = K$
$\begin{aligned} m \frac{d v}{d t} v = K \\ v d v = \frac{K}{m} d t \\ \frac{v^{2}}{2} = \frac{K}{m} t \Rightarrow v = \sqrt{\frac{2 K}{m}} t \end{aligned}$

PHXI06:WORK ENERGY AND POWER

355492 A motorcycle of mass $m$ resting on a frictionless road moves under the influence of a constant force $F$ . The work done by this force in moving the motorcycle is given by $F^{2} t^{2} / 2 m$ , where $t$ is the time in seconds.
Ratio of instantaneous power to average power of the motorcycle in $t = T$ second is:
supporting img

1

1 : 1

2

2 : 1

3

3 : 2

4

1 : 2

Explanation:

For instantaneous power
$\begin{aligned} P_{i n s} = \frac{d W}{d t} = \frac{F^{2}}{2 m} \cdot 2 t \Rightarrow P_{ins} = \frac{2 F^{2} T}{2 m} \\ P_{a v g} = \frac{Δ W}{T} = \frac{F^{2} (T^{2} - 0)}{2 m T} = \frac{F^{2} T}{2 m} \\ \frac{P_{ins}}{P_{avg}} = 2 \end{aligned}$

PHXI06:WORK ENERGY AND POWER

355488 Assertion :
Power of machine gun is determined by both, the number of bullets fired per second and kinetic energy of bullets.
Reason :
Power of any machine is defined as work done (by it) per unit time.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Kinetic energy of one bullet $= E_{k}$
Kinetic energy of $n$ bullet $= n E_{k}$
According to law of conservation of energy, the kinetic energy of bullets be equal to the work done by machine gun per sec.
Power $= \frac{work}{time} = n E_{k}$
So correct option is (1)

PHXI06:WORK ENERGY AND POWER

355489 A body is moved along a straight line by a machine delivering constant power. The distance moved by the body in time $t$ is proportional to

1

t^{2}

2

t^{3 / 2}

3

t^{1 / 2}

4

t^{3 / 4}

Explanation:

$P = F v = m a v = m (\frac{d v}{d t}) v \Rightarrow \frac{P}{m} d t = v d v$ $\Rightarrow \frac{P}{m} \times t = \frac{v^{2}}{2} \Rightarrow v = {(\frac{2 P}{m})}^{1 / 2} (t)^{1 / 2}$
Now $s = \int v d t = {(\frac{2 P}{m})}^{1 / 2} (t)^{1 / 2} d t$
$∴ s = {(\frac{2 P}{m})}^{1 / 2} [\frac{2 t^{3 / 2}}{3}] \Rightarrow s α t^{3 / 2}$

PHXI06:WORK ENERGY AND POWER

355490 A motor of power $p_{0}$ is used to deliver water at certain rate through a given horizontal pipe. To increase the rate of water through the same pipe $n$ times, the power of the motor is increased to $p_{1}$ . The ratio of $p_{1}$ to $p_{0}$ is

1

n^{4} : 1

2

n^{2} : 1

3

n : 1

4

n^{3} : 1

Explanation:

Power of motor initially $= p_{0}$
Let, rate of flow of motor $= (x)$
Since, power, $p_{0} = \frac{work}{time} = \frac{m g y}{t} = m g (\frac{y}{t})$
$\frac{y}{t} = x =$ rate of flow of water
$= m g x (1)$
If rate of flow of water is increased by $n$ times,i.e., $(n x)$
Increased power, $p_{1} = \frac{m g y^{'}}{t} = m g (\frac{y^{'}}{t})$ ,
$= nmg x$
The ratio of power
$\frac{p_{1}}{p_{0}} = \frac{n m g x}{m g x} = \frac{n}{1} \Rightarrow p_{1} : p_{0} \Rightarrow n : 1$

PHXI06:WORK ENERGY AND POWER

355491 A particle of mass $m$ is driven by a machine that delivers a constant power $K$ watts. If the particle starts from the rest the velocity of the particle at time $^{'} t^{'}$ is

1

\sqrt{\frac{2 K t}{m}}

2

\sqrt{2 K t}

3

\sqrt{\frac{m}{2 K t}}

4

\sqrt{2 K t m}

Explanation:

$p = F v = K$
$\begin{aligned} m \frac{d v}{d t} v = K \\ v d v = \frac{K}{m} d t \\ \frac{v^{2}}{2} = \frac{K}{m} t \Rightarrow v = \sqrt{\frac{2 K}{m}} t \end{aligned}$

PHXI06:WORK ENERGY AND POWER

355492 A motorcycle of mass $m$ resting on a frictionless road moves under the influence of a constant force $F$ . The work done by this force in moving the motorcycle is given by $F^{2} t^{2} / 2 m$ , where $t$ is the time in seconds.
Ratio of instantaneous power to average power of the motorcycle in $t = T$ second is:
supporting img

1

1 : 1

2

2 : 1

3

3 : 2

4

1 : 2

Explanation:

For instantaneous power
$\begin{aligned} P_{i n s} = \frac{d W}{d t} = \frac{F^{2}}{2 m} \cdot 2 t \Rightarrow P_{ins} = \frac{2 F^{2} T}{2 m} \\ P_{a v g} = \frac{Δ W}{T} = \frac{F^{2} (T^{2} - 0)}{2 m T} = \frac{F^{2} T}{2 m} \\ \frac{P_{ins}}{P_{avg}} = 2 \end{aligned}$

PHXI06:WORK ENERGY AND POWER

355488 Assertion :
Power of machine gun is determined by both, the number of bullets fired per second and kinetic energy of bullets.
Reason :
Power of any machine is defined as work done (by it) per unit time.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Kinetic energy of one bullet $= E_{k}$
Kinetic energy of $n$ bullet $= n E_{k}$
According to law of conservation of energy, the kinetic energy of bullets be equal to the work done by machine gun per sec.
Power $= \frac{work}{time} = n E_{k}$
So correct option is (1)

PHXI06:WORK ENERGY AND POWER

355489 A body is moved along a straight line by a machine delivering constant power. The distance moved by the body in time $t$ is proportional to

1

t^{2}

2

t^{3 / 2}

3

t^{1 / 2}

4

t^{3 / 4}

Explanation:

$P = F v = m a v = m (\frac{d v}{d t}) v \Rightarrow \frac{P}{m} d t = v d v$ $\Rightarrow \frac{P}{m} \times t = \frac{v^{2}}{2} \Rightarrow v = {(\frac{2 P}{m})}^{1 / 2} (t)^{1 / 2}$
Now $s = \int v d t = {(\frac{2 P}{m})}^{1 / 2} (t)^{1 / 2} d t$
$∴ s = {(\frac{2 P}{m})}^{1 / 2} [\frac{2 t^{3 / 2}}{3}] \Rightarrow s α t^{3 / 2}$

PHXI06:WORK ENERGY AND POWER

355490 A motor of power $p_{0}$ is used to deliver water at certain rate through a given horizontal pipe. To increase the rate of water through the same pipe $n$ times, the power of the motor is increased to $p_{1}$ . The ratio of $p_{1}$ to $p_{0}$ is

1

n^{4} : 1

2

n^{2} : 1

3

n : 1

4

n^{3} : 1

Explanation:

Power of motor initially $= p_{0}$
Let, rate of flow of motor $= (x)$
Since, power, $p_{0} = \frac{work}{time} = \frac{m g y}{t} = m g (\frac{y}{t})$
$\frac{y}{t} = x =$ rate of flow of water
$= m g x (1)$
If rate of flow of water is increased by $n$ times,i.e., $(n x)$
Increased power, $p_{1} = \frac{m g y^{'}}{t} = m g (\frac{y^{'}}{t})$ ,
$= nmg x$
The ratio of power
$\frac{p_{1}}{p_{0}} = \frac{n m g x}{m g x} = \frac{n}{1} \Rightarrow p_{1} : p_{0} \Rightarrow n : 1$

PHXI06:WORK ENERGY AND POWER

355491 A particle of mass $m$ is driven by a machine that delivers a constant power $K$ watts. If the particle starts from the rest the velocity of the particle at time $^{'} t^{'}$ is

1

\sqrt{\frac{2 K t}{m}}

2

\sqrt{2 K t}

3

\sqrt{\frac{m}{2 K t}}

4

\sqrt{2 K t m}

Explanation:

$p = F v = K$
$\begin{aligned} m \frac{d v}{d t} v = K \\ v d v = \frac{K}{m} d t \\ \frac{v^{2}}{2} = \frac{K}{m} t \Rightarrow v = \sqrt{\frac{2 K}{m}} t \end{aligned}$

PHXI06:WORK ENERGY AND POWER

355492 A motorcycle of mass $m$ resting on a frictionless road moves under the influence of a constant force $F$ . The work done by this force in moving the motorcycle is given by $F^{2} t^{2} / 2 m$ , where $t$ is the time in seconds.
Ratio of instantaneous power to average power of the motorcycle in $t = T$ second is:
supporting img

1

1 : 1

2

2 : 1

3

3 : 2

4

1 : 2

Explanation:

For instantaneous power
$\begin{aligned} P_{i n s} = \frac{d W}{d t} = \frac{F^{2}}{2 m} \cdot 2 t \Rightarrow P_{ins} = \frac{2 F^{2} T}{2 m} \\ P_{a v g} = \frac{Δ W}{T} = \frac{F^{2} (T^{2} - 0)}{2 m T} = \frac{F^{2} T}{2 m} \\ \frac{P_{ins}}{P_{avg}} = 2 \end{aligned}$

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