162599
An engine exerts a force \(\vec{F}=(20 \hat{i}-3 \hat{j}+5 \hat{k}) N\) and moves with velocity \(\vec{v}=(6 \hat{i}+20 \hat{j}+3 \hat{k}) \mathrm{m} / \mathrm{s}\). The power of the engine (in watt) is :
162600
An inertial frame \(S^{\prime}\) is moving with a constant velocity with respect to another inertial frame \(s\). Then
1 kinetic energy of an object when viewed from S and \(\mathrm{S}^{\prime}\) will be different
2 work done on an object when evaluated in frame \(S\) and \(S^{\prime}\) will be different
3 work energy theorem is valid in all inertial frames
4 all of these
Explanation:
NCERT-I- 74
3 RBTS PAPER
162601
A constant power \(\mathbf{P}\) is applied to a particle of mass \(\mathrm{m}\). The distance traveled by the particle when its velocity increases from \(v_1\) to \(v_2\) is (neglect friction)
1 \(\frac{m}{3 P}\left(v_2^3-v_1^3\right)\)
2 \(\frac{m}{3 P}\left(v_2-v_1\right)\)
3 \(\frac{3 P}{m}\left(v_2^2-v_1^2\right)\)
4 \(\frac{m}{3 P}\left(v_2^2-v_1^2\right)\)
Explanation:
\( P=F v \quad P=m a v \) \( P=m v \cdot \frac{d v}{d s} \cdot v \) \( \frac{p}{m} \cdot d s=v^2 \cdot d v \) Integrate both sides under limits \(V\) from \(v_1\) to \(v_2\) ds from 0 to \(S\) we get \(\left(\frac{\mathrm{v}^3}{3}\right)_{v_1}^{v_2}=\frac{P s}{m}\) \( =\frac{1}{3}\left(v_2^3-v_1^3\right)=\frac{P s}{m}=\frac{m}{3 p}\left(v_2^3-v_1^3\right) \)
NCERT-I-83
3 RBTS PAPER
162602
A stone of mass \(1 \mathrm{~kg}\) is tied to the end of a string of \(1 \mathrm{~m}\) length. It is whirled in a vertical circle. If the velocity of the stone at the top be \(4 \mathrm{~m} / \mathrm{s}\). What is the tension in the string (at that instant)?
1 \(6 \mathrm{~N}\)
2 \(16 \mathrm{~N}\)
3 \(5 \mathrm{~N}\)
4 \(10 \mathrm{~N}\)
Explanation:
At top point \( T & +M g=\frac{M v^2}{R} \) \( T & =\frac{M v^2}{R}-M g \) \( =\frac{1 \times(4)^2}{1}-1 \times 10=16-10 \Rightarrow T=6 N \)
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3 RBTS PAPER
162599
An engine exerts a force \(\vec{F}=(20 \hat{i}-3 \hat{j}+5 \hat{k}) N\) and moves with velocity \(\vec{v}=(6 \hat{i}+20 \hat{j}+3 \hat{k}) \mathrm{m} / \mathrm{s}\). The power of the engine (in watt) is :
162600
An inertial frame \(S^{\prime}\) is moving with a constant velocity with respect to another inertial frame \(s\). Then
1 kinetic energy of an object when viewed from S and \(\mathrm{S}^{\prime}\) will be different
2 work done on an object when evaluated in frame \(S\) and \(S^{\prime}\) will be different
3 work energy theorem is valid in all inertial frames
4 all of these
Explanation:
NCERT-I- 74
3 RBTS PAPER
162601
A constant power \(\mathbf{P}\) is applied to a particle of mass \(\mathrm{m}\). The distance traveled by the particle when its velocity increases from \(v_1\) to \(v_2\) is (neglect friction)
1 \(\frac{m}{3 P}\left(v_2^3-v_1^3\right)\)
2 \(\frac{m}{3 P}\left(v_2-v_1\right)\)
3 \(\frac{3 P}{m}\left(v_2^2-v_1^2\right)\)
4 \(\frac{m}{3 P}\left(v_2^2-v_1^2\right)\)
Explanation:
\( P=F v \quad P=m a v \) \( P=m v \cdot \frac{d v}{d s} \cdot v \) \( \frac{p}{m} \cdot d s=v^2 \cdot d v \) Integrate both sides under limits \(V\) from \(v_1\) to \(v_2\) ds from 0 to \(S\) we get \(\left(\frac{\mathrm{v}^3}{3}\right)_{v_1}^{v_2}=\frac{P s}{m}\) \( =\frac{1}{3}\left(v_2^3-v_1^3\right)=\frac{P s}{m}=\frac{m}{3 p}\left(v_2^3-v_1^3\right) \)
NCERT-I-83
3 RBTS PAPER
162602
A stone of mass \(1 \mathrm{~kg}\) is tied to the end of a string of \(1 \mathrm{~m}\) length. It is whirled in a vertical circle. If the velocity of the stone at the top be \(4 \mathrm{~m} / \mathrm{s}\). What is the tension in the string (at that instant)?
1 \(6 \mathrm{~N}\)
2 \(16 \mathrm{~N}\)
3 \(5 \mathrm{~N}\)
4 \(10 \mathrm{~N}\)
Explanation:
At top point \( T & +M g=\frac{M v^2}{R} \) \( T & =\frac{M v^2}{R}-M g \) \( =\frac{1 \times(4)^2}{1}-1 \times 10=16-10 \Rightarrow T=6 N \)
162599
An engine exerts a force \(\vec{F}=(20 \hat{i}-3 \hat{j}+5 \hat{k}) N\) and moves with velocity \(\vec{v}=(6 \hat{i}+20 \hat{j}+3 \hat{k}) \mathrm{m} / \mathrm{s}\). The power of the engine (in watt) is :
162600
An inertial frame \(S^{\prime}\) is moving with a constant velocity with respect to another inertial frame \(s\). Then
1 kinetic energy of an object when viewed from S and \(\mathrm{S}^{\prime}\) will be different
2 work done on an object when evaluated in frame \(S\) and \(S^{\prime}\) will be different
3 work energy theorem is valid in all inertial frames
4 all of these
Explanation:
NCERT-I- 74
3 RBTS PAPER
162601
A constant power \(\mathbf{P}\) is applied to a particle of mass \(\mathrm{m}\). The distance traveled by the particle when its velocity increases from \(v_1\) to \(v_2\) is (neglect friction)
1 \(\frac{m}{3 P}\left(v_2^3-v_1^3\right)\)
2 \(\frac{m}{3 P}\left(v_2-v_1\right)\)
3 \(\frac{3 P}{m}\left(v_2^2-v_1^2\right)\)
4 \(\frac{m}{3 P}\left(v_2^2-v_1^2\right)\)
Explanation:
\( P=F v \quad P=m a v \) \( P=m v \cdot \frac{d v}{d s} \cdot v \) \( \frac{p}{m} \cdot d s=v^2 \cdot d v \) Integrate both sides under limits \(V\) from \(v_1\) to \(v_2\) ds from 0 to \(S\) we get \(\left(\frac{\mathrm{v}^3}{3}\right)_{v_1}^{v_2}=\frac{P s}{m}\) \( =\frac{1}{3}\left(v_2^3-v_1^3\right)=\frac{P s}{m}=\frac{m}{3 p}\left(v_2^3-v_1^3\right) \)
NCERT-I-83
3 RBTS PAPER
162602
A stone of mass \(1 \mathrm{~kg}\) is tied to the end of a string of \(1 \mathrm{~m}\) length. It is whirled in a vertical circle. If the velocity of the stone at the top be \(4 \mathrm{~m} / \mathrm{s}\). What is the tension in the string (at that instant)?
1 \(6 \mathrm{~N}\)
2 \(16 \mathrm{~N}\)
3 \(5 \mathrm{~N}\)
4 \(10 \mathrm{~N}\)
Explanation:
At top point \( T & +M g=\frac{M v^2}{R} \) \( T & =\frac{M v^2}{R}-M g \) \( =\frac{1 \times(4)^2}{1}-1 \times 10=16-10 \Rightarrow T=6 N \)
162599
An engine exerts a force \(\vec{F}=(20 \hat{i}-3 \hat{j}+5 \hat{k}) N\) and moves with velocity \(\vec{v}=(6 \hat{i}+20 \hat{j}+3 \hat{k}) \mathrm{m} / \mathrm{s}\). The power of the engine (in watt) is :
162600
An inertial frame \(S^{\prime}\) is moving with a constant velocity with respect to another inertial frame \(s\). Then
1 kinetic energy of an object when viewed from S and \(\mathrm{S}^{\prime}\) will be different
2 work done on an object when evaluated in frame \(S\) and \(S^{\prime}\) will be different
3 work energy theorem is valid in all inertial frames
4 all of these
Explanation:
NCERT-I- 74
3 RBTS PAPER
162601
A constant power \(\mathbf{P}\) is applied to a particle of mass \(\mathrm{m}\). The distance traveled by the particle when its velocity increases from \(v_1\) to \(v_2\) is (neglect friction)
1 \(\frac{m}{3 P}\left(v_2^3-v_1^3\right)\)
2 \(\frac{m}{3 P}\left(v_2-v_1\right)\)
3 \(\frac{3 P}{m}\left(v_2^2-v_1^2\right)\)
4 \(\frac{m}{3 P}\left(v_2^2-v_1^2\right)\)
Explanation:
\( P=F v \quad P=m a v \) \( P=m v \cdot \frac{d v}{d s} \cdot v \) \( \frac{p}{m} \cdot d s=v^2 \cdot d v \) Integrate both sides under limits \(V\) from \(v_1\) to \(v_2\) ds from 0 to \(S\) we get \(\left(\frac{\mathrm{v}^3}{3}\right)_{v_1}^{v_2}=\frac{P s}{m}\) \( =\frac{1}{3}\left(v_2^3-v_1^3\right)=\frac{P s}{m}=\frac{m}{3 p}\left(v_2^3-v_1^3\right) \)
NCERT-I-83
3 RBTS PAPER
162602
A stone of mass \(1 \mathrm{~kg}\) is tied to the end of a string of \(1 \mathrm{~m}\) length. It is whirled in a vertical circle. If the velocity of the stone at the top be \(4 \mathrm{~m} / \mathrm{s}\). What is the tension in the string (at that instant)?
1 \(6 \mathrm{~N}\)
2 \(16 \mathrm{~N}\)
3 \(5 \mathrm{~N}\)
4 \(10 \mathrm{~N}\)
Explanation:
At top point \( T & +M g=\frac{M v^2}{R} \) \( T & =\frac{M v^2}{R}-M g \) \( =\frac{1 \times(4)^2}{1}-1 \times 10=16-10 \Rightarrow T=6 N \)