162575
A particle of mass \(m\) is driven by a machine that delivers a constant power \(\mathbf{k}\) watts. If the particle starts from rest the force on the particle at time \(t\) is:
1 \(\sqrt{\frac{m k}{2}} t^{-1 / 2}\)
2 \(\sqrt{m k} t^{-1 / 2}\)
3 \(\sqrt{2 m k} t^{-1 / 2}\)
4 \(\frac{1}{2} \sqrt{m k} t^{-1 / 2}\)
Explanation:
Power \(=\mathrm{K}\) we know that \(P=F . V\) \( K=F \cdot V \) \( K=m \frac{d V}{d t} \cdot V \) \( \frac{K}{m} \int d t=\int V d V \) \( \frac{K}{m} t=\frac{V^2}{2} \) \( V=\sqrt{\frac{2 K}{m}} t^{1 / 2} \) \( a=\frac{d V}{d t}=\sqrt{\frac{2 K}{m}} \frac{t^{-1 / 2}}{2} \) \( F=m a \) \( F=\sqrt{\frac{m k}{2}} t^{-1 / 2} . \)
NCERT-I-83
3 RBTS PAPER
162576
A man does a given amount of work in \(10 \mathrm{sec}\). Another man does the same amount of work in \(20 \mathrm{sec}\). The ratio of the output power of first man to the second man is :
1 1
2 \(1 / 2\)
3 \(2 / 1\)
4 None of these
Explanation:
Power \(=\frac{W}{t}\). If \(W\) is constant then \(P \propto \frac{1}{t}\) i.e., \(\frac{P_1}{P_2}=\frac{t_2}{t_1}=\frac{20}{10}=\frac{2}{1}\)
NCERT-I-83
3 RBTS PAPER
162577
Power supplied to a body of mass \(2 \mathrm{~kg}\) varies with time as \(P=\frac{3 t^2}{2}\) watt . Here \(t\) is in seconds . If velocity of particle at \(t=0\) is \(v=0\), the velocity of particle at time \(\mathbf{t}=\mathbf{2} \mathrm{s}\) will be :
1 \(1 \mathrm{~m} / \mathrm{s}\)
2 \(4 \mathrm{~m} / \mathrm{s}\)
3 \(2 \mathrm{~m} / \mathrm{s}\)
4 \(2 \sqrt{3} \mathrm{~m} / \mathrm{s}\)
Explanation:
We know that , \( W=\int_0^2 P d t \) \( W=\int_0^2 \frac{3 t^2}{2} d t \) \( W=\frac{1}{2}\left(2^3\right) \) \( \frac{1}{2} m v^2=\frac{1}{2} \times 8 \Rightarrow 2 \times v^2=8 \Rightarrow v=2 \mathrm{~m} / \mathrm{s} \)
NCERT-I-83
3 RBTS PAPER
162578
If the momentum of a body increases by \(20 \%\), the percentage increase in its K.E. is equal to :
1 44
2 66
3 20
4 88
Explanation:
Momentum increases by \(20 \%\) means if momentum were 100 units, it becomes 120, thereby increasing velocity \(\frac{120}{100}=1.2\) times. KE increaes to \((1.2)^2\) \(=1.44\) times. so KE increases from 100 to 144 . Therefore, the percentage increaes of KE is \(44 \%\).
NCERT-I-75
3 RBTS PAPER
162579
A body is moving unidirectionally under the influence of a source of constant power. Its displacement in time \(t\) is proportional to
162575
A particle of mass \(m\) is driven by a machine that delivers a constant power \(\mathbf{k}\) watts. If the particle starts from rest the force on the particle at time \(t\) is:
1 \(\sqrt{\frac{m k}{2}} t^{-1 / 2}\)
2 \(\sqrt{m k} t^{-1 / 2}\)
3 \(\sqrt{2 m k} t^{-1 / 2}\)
4 \(\frac{1}{2} \sqrt{m k} t^{-1 / 2}\)
Explanation:
Power \(=\mathrm{K}\) we know that \(P=F . V\) \( K=F \cdot V \) \( K=m \frac{d V}{d t} \cdot V \) \( \frac{K}{m} \int d t=\int V d V \) \( \frac{K}{m} t=\frac{V^2}{2} \) \( V=\sqrt{\frac{2 K}{m}} t^{1 / 2} \) \( a=\frac{d V}{d t}=\sqrt{\frac{2 K}{m}} \frac{t^{-1 / 2}}{2} \) \( F=m a \) \( F=\sqrt{\frac{m k}{2}} t^{-1 / 2} . \)
NCERT-I-83
3 RBTS PAPER
162576
A man does a given amount of work in \(10 \mathrm{sec}\). Another man does the same amount of work in \(20 \mathrm{sec}\). The ratio of the output power of first man to the second man is :
1 1
2 \(1 / 2\)
3 \(2 / 1\)
4 None of these
Explanation:
Power \(=\frac{W}{t}\). If \(W\) is constant then \(P \propto \frac{1}{t}\) i.e., \(\frac{P_1}{P_2}=\frac{t_2}{t_1}=\frac{20}{10}=\frac{2}{1}\)
NCERT-I-83
3 RBTS PAPER
162577
Power supplied to a body of mass \(2 \mathrm{~kg}\) varies with time as \(P=\frac{3 t^2}{2}\) watt . Here \(t\) is in seconds . If velocity of particle at \(t=0\) is \(v=0\), the velocity of particle at time \(\mathbf{t}=\mathbf{2} \mathrm{s}\) will be :
1 \(1 \mathrm{~m} / \mathrm{s}\)
2 \(4 \mathrm{~m} / \mathrm{s}\)
3 \(2 \mathrm{~m} / \mathrm{s}\)
4 \(2 \sqrt{3} \mathrm{~m} / \mathrm{s}\)
Explanation:
We know that , \( W=\int_0^2 P d t \) \( W=\int_0^2 \frac{3 t^2}{2} d t \) \( W=\frac{1}{2}\left(2^3\right) \) \( \frac{1}{2} m v^2=\frac{1}{2} \times 8 \Rightarrow 2 \times v^2=8 \Rightarrow v=2 \mathrm{~m} / \mathrm{s} \)
NCERT-I-83
3 RBTS PAPER
162578
If the momentum of a body increases by \(20 \%\), the percentage increase in its K.E. is equal to :
1 44
2 66
3 20
4 88
Explanation:
Momentum increases by \(20 \%\) means if momentum were 100 units, it becomes 120, thereby increasing velocity \(\frac{120}{100}=1.2\) times. KE increaes to \((1.2)^2\) \(=1.44\) times. so KE increases from 100 to 144 . Therefore, the percentage increaes of KE is \(44 \%\).
NCERT-I-75
3 RBTS PAPER
162579
A body is moving unidirectionally under the influence of a source of constant power. Its displacement in time \(t\) is proportional to
162575
A particle of mass \(m\) is driven by a machine that delivers a constant power \(\mathbf{k}\) watts. If the particle starts from rest the force on the particle at time \(t\) is:
1 \(\sqrt{\frac{m k}{2}} t^{-1 / 2}\)
2 \(\sqrt{m k} t^{-1 / 2}\)
3 \(\sqrt{2 m k} t^{-1 / 2}\)
4 \(\frac{1}{2} \sqrt{m k} t^{-1 / 2}\)
Explanation:
Power \(=\mathrm{K}\) we know that \(P=F . V\) \( K=F \cdot V \) \( K=m \frac{d V}{d t} \cdot V \) \( \frac{K}{m} \int d t=\int V d V \) \( \frac{K}{m} t=\frac{V^2}{2} \) \( V=\sqrt{\frac{2 K}{m}} t^{1 / 2} \) \( a=\frac{d V}{d t}=\sqrt{\frac{2 K}{m}} \frac{t^{-1 / 2}}{2} \) \( F=m a \) \( F=\sqrt{\frac{m k}{2}} t^{-1 / 2} . \)
NCERT-I-83
3 RBTS PAPER
162576
A man does a given amount of work in \(10 \mathrm{sec}\). Another man does the same amount of work in \(20 \mathrm{sec}\). The ratio of the output power of first man to the second man is :
1 1
2 \(1 / 2\)
3 \(2 / 1\)
4 None of these
Explanation:
Power \(=\frac{W}{t}\). If \(W\) is constant then \(P \propto \frac{1}{t}\) i.e., \(\frac{P_1}{P_2}=\frac{t_2}{t_1}=\frac{20}{10}=\frac{2}{1}\)
NCERT-I-83
3 RBTS PAPER
162577
Power supplied to a body of mass \(2 \mathrm{~kg}\) varies with time as \(P=\frac{3 t^2}{2}\) watt . Here \(t\) is in seconds . If velocity of particle at \(t=0\) is \(v=0\), the velocity of particle at time \(\mathbf{t}=\mathbf{2} \mathrm{s}\) will be :
1 \(1 \mathrm{~m} / \mathrm{s}\)
2 \(4 \mathrm{~m} / \mathrm{s}\)
3 \(2 \mathrm{~m} / \mathrm{s}\)
4 \(2 \sqrt{3} \mathrm{~m} / \mathrm{s}\)
Explanation:
We know that , \( W=\int_0^2 P d t \) \( W=\int_0^2 \frac{3 t^2}{2} d t \) \( W=\frac{1}{2}\left(2^3\right) \) \( \frac{1}{2} m v^2=\frac{1}{2} \times 8 \Rightarrow 2 \times v^2=8 \Rightarrow v=2 \mathrm{~m} / \mathrm{s} \)
NCERT-I-83
3 RBTS PAPER
162578
If the momentum of a body increases by \(20 \%\), the percentage increase in its K.E. is equal to :
1 44
2 66
3 20
4 88
Explanation:
Momentum increases by \(20 \%\) means if momentum were 100 units, it becomes 120, thereby increasing velocity \(\frac{120}{100}=1.2\) times. KE increaes to \((1.2)^2\) \(=1.44\) times. so KE increases from 100 to 144 . Therefore, the percentage increaes of KE is \(44 \%\).
NCERT-I-75
3 RBTS PAPER
162579
A body is moving unidirectionally under the influence of a source of constant power. Its displacement in time \(t\) is proportional to
162575
A particle of mass \(m\) is driven by a machine that delivers a constant power \(\mathbf{k}\) watts. If the particle starts from rest the force on the particle at time \(t\) is:
1 \(\sqrt{\frac{m k}{2}} t^{-1 / 2}\)
2 \(\sqrt{m k} t^{-1 / 2}\)
3 \(\sqrt{2 m k} t^{-1 / 2}\)
4 \(\frac{1}{2} \sqrt{m k} t^{-1 / 2}\)
Explanation:
Power \(=\mathrm{K}\) we know that \(P=F . V\) \( K=F \cdot V \) \( K=m \frac{d V}{d t} \cdot V \) \( \frac{K}{m} \int d t=\int V d V \) \( \frac{K}{m} t=\frac{V^2}{2} \) \( V=\sqrt{\frac{2 K}{m}} t^{1 / 2} \) \( a=\frac{d V}{d t}=\sqrt{\frac{2 K}{m}} \frac{t^{-1 / 2}}{2} \) \( F=m a \) \( F=\sqrt{\frac{m k}{2}} t^{-1 / 2} . \)
NCERT-I-83
3 RBTS PAPER
162576
A man does a given amount of work in \(10 \mathrm{sec}\). Another man does the same amount of work in \(20 \mathrm{sec}\). The ratio of the output power of first man to the second man is :
1 1
2 \(1 / 2\)
3 \(2 / 1\)
4 None of these
Explanation:
Power \(=\frac{W}{t}\). If \(W\) is constant then \(P \propto \frac{1}{t}\) i.e., \(\frac{P_1}{P_2}=\frac{t_2}{t_1}=\frac{20}{10}=\frac{2}{1}\)
NCERT-I-83
3 RBTS PAPER
162577
Power supplied to a body of mass \(2 \mathrm{~kg}\) varies with time as \(P=\frac{3 t^2}{2}\) watt . Here \(t\) is in seconds . If velocity of particle at \(t=0\) is \(v=0\), the velocity of particle at time \(\mathbf{t}=\mathbf{2} \mathrm{s}\) will be :
1 \(1 \mathrm{~m} / \mathrm{s}\)
2 \(4 \mathrm{~m} / \mathrm{s}\)
3 \(2 \mathrm{~m} / \mathrm{s}\)
4 \(2 \sqrt{3} \mathrm{~m} / \mathrm{s}\)
Explanation:
We know that , \( W=\int_0^2 P d t \) \( W=\int_0^2 \frac{3 t^2}{2} d t \) \( W=\frac{1}{2}\left(2^3\right) \) \( \frac{1}{2} m v^2=\frac{1}{2} \times 8 \Rightarrow 2 \times v^2=8 \Rightarrow v=2 \mathrm{~m} / \mathrm{s} \)
NCERT-I-83
3 RBTS PAPER
162578
If the momentum of a body increases by \(20 \%\), the percentage increase in its K.E. is equal to :
1 44
2 66
3 20
4 88
Explanation:
Momentum increases by \(20 \%\) means if momentum were 100 units, it becomes 120, thereby increasing velocity \(\frac{120}{100}=1.2\) times. KE increaes to \((1.2)^2\) \(=1.44\) times. so KE increases from 100 to 144 . Therefore, the percentage increaes of KE is \(44 \%\).
NCERT-I-75
3 RBTS PAPER
162579
A body is moving unidirectionally under the influence of a source of constant power. Its displacement in time \(t\) is proportional to
162575
A particle of mass \(m\) is driven by a machine that delivers a constant power \(\mathbf{k}\) watts. If the particle starts from rest the force on the particle at time \(t\) is:
1 \(\sqrt{\frac{m k}{2}} t^{-1 / 2}\)
2 \(\sqrt{m k} t^{-1 / 2}\)
3 \(\sqrt{2 m k} t^{-1 / 2}\)
4 \(\frac{1}{2} \sqrt{m k} t^{-1 / 2}\)
Explanation:
Power \(=\mathrm{K}\) we know that \(P=F . V\) \( K=F \cdot V \) \( K=m \frac{d V}{d t} \cdot V \) \( \frac{K}{m} \int d t=\int V d V \) \( \frac{K}{m} t=\frac{V^2}{2} \) \( V=\sqrt{\frac{2 K}{m}} t^{1 / 2} \) \( a=\frac{d V}{d t}=\sqrt{\frac{2 K}{m}} \frac{t^{-1 / 2}}{2} \) \( F=m a \) \( F=\sqrt{\frac{m k}{2}} t^{-1 / 2} . \)
NCERT-I-83
3 RBTS PAPER
162576
A man does a given amount of work in \(10 \mathrm{sec}\). Another man does the same amount of work in \(20 \mathrm{sec}\). The ratio of the output power of first man to the second man is :
1 1
2 \(1 / 2\)
3 \(2 / 1\)
4 None of these
Explanation:
Power \(=\frac{W}{t}\). If \(W\) is constant then \(P \propto \frac{1}{t}\) i.e., \(\frac{P_1}{P_2}=\frac{t_2}{t_1}=\frac{20}{10}=\frac{2}{1}\)
NCERT-I-83
3 RBTS PAPER
162577
Power supplied to a body of mass \(2 \mathrm{~kg}\) varies with time as \(P=\frac{3 t^2}{2}\) watt . Here \(t\) is in seconds . If velocity of particle at \(t=0\) is \(v=0\), the velocity of particle at time \(\mathbf{t}=\mathbf{2} \mathrm{s}\) will be :
1 \(1 \mathrm{~m} / \mathrm{s}\)
2 \(4 \mathrm{~m} / \mathrm{s}\)
3 \(2 \mathrm{~m} / \mathrm{s}\)
4 \(2 \sqrt{3} \mathrm{~m} / \mathrm{s}\)
Explanation:
We know that , \( W=\int_0^2 P d t \) \( W=\int_0^2 \frac{3 t^2}{2} d t \) \( W=\frac{1}{2}\left(2^3\right) \) \( \frac{1}{2} m v^2=\frac{1}{2} \times 8 \Rightarrow 2 \times v^2=8 \Rightarrow v=2 \mathrm{~m} / \mathrm{s} \)
NCERT-I-83
3 RBTS PAPER
162578
If the momentum of a body increases by \(20 \%\), the percentage increase in its K.E. is equal to :
1 44
2 66
3 20
4 88
Explanation:
Momentum increases by \(20 \%\) means if momentum were 100 units, it becomes 120, thereby increasing velocity \(\frac{120}{100}=1.2\) times. KE increaes to \((1.2)^2\) \(=1.44\) times. so KE increases from 100 to 144 . Therefore, the percentage increaes of KE is \(44 \%\).
NCERT-I-75
3 RBTS PAPER
162579
A body is moving unidirectionally under the influence of a source of constant power. Its displacement in time \(t\) is proportional to
