NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI06:WORK ENERGY AND POWER
355497
Assertion : In one dimensional motion with constant acceleration, the power delivered is proportional to time. Reason : In general, power can be expressed as dot product of force applied and displacement.
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:
Power \(P=\dfrac{W}{t}=\dfrac{\vec{F} \cdot \vec{s}}{t}=\vec{F} \cdot \vec{v}=(m a) \times a t=m a^{2} t\) \(\therefore P \propto t\) Thus, power is proportional to time. So Assertion is correct and Reason is wrong. So correct option is (3).
PHXI06:WORK ENERGY AND POWER
355498
A body is moving unidirectionally under the influence of a constant power source. Its displacement in time \(t\) is proportional to
1 \(t^{2}\)
2 \(t^{3 / 2}\)
3 \(t^{2 / 3}\)
4 \(t\)
Explanation:
Let the constant power be \(P\). Here, initial velocity, \(u=0\) Instantaneous power of a body, \(P=F v\) \(\therefore \quad P = mav\) \(\begin{aligned}& P=m\left(\dfrac{d v}{d t}\right) v \\& \Rightarrow v d v=\dfrac{P}{m} d t \\& \int_{0}^{v} d v=\dfrac{P}{m} \int_{0}^{t} d t \\& \Rightarrow \dfrac{v^{2}}{2}=\dfrac{P}{m} t \\& v=\sqrt{\dfrac{2 P t}{m}}=\sqrt{\dfrac{2 P}{m}} t^{1 / 2} \\& v=\dfrac{d S}{d t}=\sqrt{\dfrac{2 P}{m}} t^{1 / 2} \\& \Rightarrow d S=\sqrt{\dfrac{2 P}{m}} t^{1 / 2} d t \\& \int d S=\int \sqrt{\dfrac{2 P}{m}} t^{1 / 2} d t \\& S=\dfrac{2}{3} \sqrt{\dfrac{2 P}{m}} t^{3 / 2} \\& \therefore S \propto t^{3 / 2}\end{aligned}\)
JEE - 2024
PHXI06:WORK ENERGY AND POWER
355499
A particle moves with a velocity \((5\hat i + 3\hat j + 6\hat k)m{\rm{/}}s\) under the influence of a constant force \((10 \hat{i}+10 \hat{j}+20 \hat{k}) N\). The instantaneous power applied to the particle is
355500
An elevator that can carry a maximum load of 1800\(kg\) (elevator + passengers) is moving up with a constant speed of \(2\;m{s^{ - 1}}\). The frictional force opposing the motion is 400 \(N\). What is minimum power delivered by the motor to the elevator?
1 88 \(kW\)
2 22 \(kW\)
3 44 \(kW\)
4 66 \(kW\)
Explanation:
Here, \(m = 1800\,kg\) Frictional force, \(f = 4000N\) \(v = 2\,m{s^{ - 1}}\) Downward force on elevator is \(F = mg + f\) \( = \left( {1800kg \times 10m{s^{ - 2}}} \right) + 4000N = 22000N\) The motor must supply enough power to balance this force. Hence, \(P = Fv = \left( {22000N} \right)\left( {2m{s^{ - 1}}} \right) = 44\,kW\)
355497
Assertion : In one dimensional motion with constant acceleration, the power delivered is proportional to time. Reason : In general, power can be expressed as dot product of force applied and displacement.
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:
Power \(P=\dfrac{W}{t}=\dfrac{\vec{F} \cdot \vec{s}}{t}=\vec{F} \cdot \vec{v}=(m a) \times a t=m a^{2} t\) \(\therefore P \propto t\) Thus, power is proportional to time. So Assertion is correct and Reason is wrong. So correct option is (3).
PHXI06:WORK ENERGY AND POWER
355498
A body is moving unidirectionally under the influence of a constant power source. Its displacement in time \(t\) is proportional to
1 \(t^{2}\)
2 \(t^{3 / 2}\)
3 \(t^{2 / 3}\)
4 \(t\)
Explanation:
Let the constant power be \(P\). Here, initial velocity, \(u=0\) Instantaneous power of a body, \(P=F v\) \(\therefore \quad P = mav\) \(\begin{aligned}& P=m\left(\dfrac{d v}{d t}\right) v \\& \Rightarrow v d v=\dfrac{P}{m} d t \\& \int_{0}^{v} d v=\dfrac{P}{m} \int_{0}^{t} d t \\& \Rightarrow \dfrac{v^{2}}{2}=\dfrac{P}{m} t \\& v=\sqrt{\dfrac{2 P t}{m}}=\sqrt{\dfrac{2 P}{m}} t^{1 / 2} \\& v=\dfrac{d S}{d t}=\sqrt{\dfrac{2 P}{m}} t^{1 / 2} \\& \Rightarrow d S=\sqrt{\dfrac{2 P}{m}} t^{1 / 2} d t \\& \int d S=\int \sqrt{\dfrac{2 P}{m}} t^{1 / 2} d t \\& S=\dfrac{2}{3} \sqrt{\dfrac{2 P}{m}} t^{3 / 2} \\& \therefore S \propto t^{3 / 2}\end{aligned}\)
JEE - 2024
PHXI06:WORK ENERGY AND POWER
355499
A particle moves with a velocity \((5\hat i + 3\hat j + 6\hat k)m{\rm{/}}s\) under the influence of a constant force \((10 \hat{i}+10 \hat{j}+20 \hat{k}) N\). The instantaneous power applied to the particle is
355500
An elevator that can carry a maximum load of 1800\(kg\) (elevator + passengers) is moving up with a constant speed of \(2\;m{s^{ - 1}}\). The frictional force opposing the motion is 400 \(N\). What is minimum power delivered by the motor to the elevator?
1 88 \(kW\)
2 22 \(kW\)
3 44 \(kW\)
4 66 \(kW\)
Explanation:
Here, \(m = 1800\,kg\) Frictional force, \(f = 4000N\) \(v = 2\,m{s^{ - 1}}\) Downward force on elevator is \(F = mg + f\) \( = \left( {1800kg \times 10m{s^{ - 2}}} \right) + 4000N = 22000N\) The motor must supply enough power to balance this force. Hence, \(P = Fv = \left( {22000N} \right)\left( {2m{s^{ - 1}}} \right) = 44\,kW\)
355497
Assertion : In one dimensional motion with constant acceleration, the power delivered is proportional to time. Reason : In general, power can be expressed as dot product of force applied and displacement.
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:
Power \(P=\dfrac{W}{t}=\dfrac{\vec{F} \cdot \vec{s}}{t}=\vec{F} \cdot \vec{v}=(m a) \times a t=m a^{2} t\) \(\therefore P \propto t\) Thus, power is proportional to time. So Assertion is correct and Reason is wrong. So correct option is (3).
PHXI06:WORK ENERGY AND POWER
355498
A body is moving unidirectionally under the influence of a constant power source. Its displacement in time \(t\) is proportional to
1 \(t^{2}\)
2 \(t^{3 / 2}\)
3 \(t^{2 / 3}\)
4 \(t\)
Explanation:
Let the constant power be \(P\). Here, initial velocity, \(u=0\) Instantaneous power of a body, \(P=F v\) \(\therefore \quad P = mav\) \(\begin{aligned}& P=m\left(\dfrac{d v}{d t}\right) v \\& \Rightarrow v d v=\dfrac{P}{m} d t \\& \int_{0}^{v} d v=\dfrac{P}{m} \int_{0}^{t} d t \\& \Rightarrow \dfrac{v^{2}}{2}=\dfrac{P}{m} t \\& v=\sqrt{\dfrac{2 P t}{m}}=\sqrt{\dfrac{2 P}{m}} t^{1 / 2} \\& v=\dfrac{d S}{d t}=\sqrt{\dfrac{2 P}{m}} t^{1 / 2} \\& \Rightarrow d S=\sqrt{\dfrac{2 P}{m}} t^{1 / 2} d t \\& \int d S=\int \sqrt{\dfrac{2 P}{m}} t^{1 / 2} d t \\& S=\dfrac{2}{3} \sqrt{\dfrac{2 P}{m}} t^{3 / 2} \\& \therefore S \propto t^{3 / 2}\end{aligned}\)
JEE - 2024
PHXI06:WORK ENERGY AND POWER
355499
A particle moves with a velocity \((5\hat i + 3\hat j + 6\hat k)m{\rm{/}}s\) under the influence of a constant force \((10 \hat{i}+10 \hat{j}+20 \hat{k}) N\). The instantaneous power applied to the particle is
355500
An elevator that can carry a maximum load of 1800\(kg\) (elevator + passengers) is moving up with a constant speed of \(2\;m{s^{ - 1}}\). The frictional force opposing the motion is 400 \(N\). What is minimum power delivered by the motor to the elevator?
1 88 \(kW\)
2 22 \(kW\)
3 44 \(kW\)
4 66 \(kW\)
Explanation:
Here, \(m = 1800\,kg\) Frictional force, \(f = 4000N\) \(v = 2\,m{s^{ - 1}}\) Downward force on elevator is \(F = mg + f\) \( = \left( {1800kg \times 10m{s^{ - 2}}} \right) + 4000N = 22000N\) The motor must supply enough power to balance this force. Hence, \(P = Fv = \left( {22000N} \right)\left( {2m{s^{ - 1}}} \right) = 44\,kW\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
PHXI06:WORK ENERGY AND POWER
355497
Assertion : In one dimensional motion with constant acceleration, the power delivered is proportional to time. Reason : In general, power can be expressed as dot product of force applied and displacement.
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:
Power \(P=\dfrac{W}{t}=\dfrac{\vec{F} \cdot \vec{s}}{t}=\vec{F} \cdot \vec{v}=(m a) \times a t=m a^{2} t\) \(\therefore P \propto t\) Thus, power is proportional to time. So Assertion is correct and Reason is wrong. So correct option is (3).
PHXI06:WORK ENERGY AND POWER
355498
A body is moving unidirectionally under the influence of a constant power source. Its displacement in time \(t\) is proportional to
1 \(t^{2}\)
2 \(t^{3 / 2}\)
3 \(t^{2 / 3}\)
4 \(t\)
Explanation:
Let the constant power be \(P\). Here, initial velocity, \(u=0\) Instantaneous power of a body, \(P=F v\) \(\therefore \quad P = mav\) \(\begin{aligned}& P=m\left(\dfrac{d v}{d t}\right) v \\& \Rightarrow v d v=\dfrac{P}{m} d t \\& \int_{0}^{v} d v=\dfrac{P}{m} \int_{0}^{t} d t \\& \Rightarrow \dfrac{v^{2}}{2}=\dfrac{P}{m} t \\& v=\sqrt{\dfrac{2 P t}{m}}=\sqrt{\dfrac{2 P}{m}} t^{1 / 2} \\& v=\dfrac{d S}{d t}=\sqrt{\dfrac{2 P}{m}} t^{1 / 2} \\& \Rightarrow d S=\sqrt{\dfrac{2 P}{m}} t^{1 / 2} d t \\& \int d S=\int \sqrt{\dfrac{2 P}{m}} t^{1 / 2} d t \\& S=\dfrac{2}{3} \sqrt{\dfrac{2 P}{m}} t^{3 / 2} \\& \therefore S \propto t^{3 / 2}\end{aligned}\)
JEE - 2024
PHXI06:WORK ENERGY AND POWER
355499
A particle moves with a velocity \((5\hat i + 3\hat j + 6\hat k)m{\rm{/}}s\) under the influence of a constant force \((10 \hat{i}+10 \hat{j}+20 \hat{k}) N\). The instantaneous power applied to the particle is
355500
An elevator that can carry a maximum load of 1800\(kg\) (elevator + passengers) is moving up with a constant speed of \(2\;m{s^{ - 1}}\). The frictional force opposing the motion is 400 \(N\). What is minimum power delivered by the motor to the elevator?
1 88 \(kW\)
2 22 \(kW\)
3 44 \(kW\)
4 66 \(kW\)
Explanation:
Here, \(m = 1800\,kg\) Frictional force, \(f = 4000N\) \(v = 2\,m{s^{ - 1}}\) Downward force on elevator is \(F = mg + f\) \( = \left( {1800kg \times 10m{s^{ - 2}}} \right) + 4000N = 22000N\) The motor must supply enough power to balance this force. Hence, \(P = Fv = \left( {22000N} \right)\left( {2m{s^{ - 1}}} \right) = 44\,kW\)