NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
366179
In an orbital motion, the angular momentum vector is:
1 Parallel to the linear momentum
2 Along the radius vector
3 Perpendicular to the orbital plane
4 In the orbital plane
Explanation:
Option (3) is correct
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
366180
Assertion : A particle moving on a straight line with a uniform velocity, its angular momentum is always zero. Reason : The angular momentum is a vector.
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:
\(|\vec{L}|=m v r \sin \theta\) If straight line, \(\theta=0\) \(\Rightarrow L=0\). Angular momentum is zero only if the particle's velocity vector is parallel to the position vector so assertion is wrong. The angular momentum is a vector quantity. So correct option is (4).
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
366181
A particle moves in a straight line with constant acceleration. Its angular momentum about any fixed point is :
1 constant
2 directly proportional to time
3 inversely proportional to time
4 inversely proportional to distance covered
Explanation:
Let a particle \({A}\) of mass \({m}\) is moving in a straight line with velocity \({v}\) as shown below. Angular momentum of particle about any fixed point say \({O}\) will be \(\begin{aligned}& L=m\left(\overrightarrow{O A}^{\prime} \times \vec{v}\right) \\& L=m v O A^{\prime} \sin \theta=m v r \\& \Rightarrow L=m(a t) r \quad[a=\text { constant acceleration }] \\& L=(m a r) t \\& \therefore L \propto t\end{aligned}\). So correct option is (2)
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
366182
Statement A : Angular momentum of a body depends on axis of measurement. Statement B : Torque acting on a body depends on axis of measurement
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both statements are correct.
4 Both Statements are incorrect.
Explanation:
\(\bar{L}=\bar{r} \times \bar{P}, \vec{\tau}=\bar{r} \times \bar{P}\) So both \(\bar{L}\) and \(\vec{\tau}\) depend on axis about which they are measured.
366179
In an orbital motion, the angular momentum vector is:
1 Parallel to the linear momentum
2 Along the radius vector
3 Perpendicular to the orbital plane
4 In the orbital plane
Explanation:
Option (3) is correct
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
366180
Assertion : A particle moving on a straight line with a uniform velocity, its angular momentum is always zero. Reason : The angular momentum is a vector.
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:
\(|\vec{L}|=m v r \sin \theta\) If straight line, \(\theta=0\) \(\Rightarrow L=0\). Angular momentum is zero only if the particle's velocity vector is parallel to the position vector so assertion is wrong. The angular momentum is a vector quantity. So correct option is (4).
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
366181
A particle moves in a straight line with constant acceleration. Its angular momentum about any fixed point is :
1 constant
2 directly proportional to time
3 inversely proportional to time
4 inversely proportional to distance covered
Explanation:
Let a particle \({A}\) of mass \({m}\) is moving in a straight line with velocity \({v}\) as shown below. Angular momentum of particle about any fixed point say \({O}\) will be \(\begin{aligned}& L=m\left(\overrightarrow{O A}^{\prime} \times \vec{v}\right) \\& L=m v O A^{\prime} \sin \theta=m v r \\& \Rightarrow L=m(a t) r \quad[a=\text { constant acceleration }] \\& L=(m a r) t \\& \therefore L \propto t\end{aligned}\). So correct option is (2)
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
366182
Statement A : Angular momentum of a body depends on axis of measurement. Statement B : Torque acting on a body depends on axis of measurement
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both statements are correct.
4 Both Statements are incorrect.
Explanation:
\(\bar{L}=\bar{r} \times \bar{P}, \vec{\tau}=\bar{r} \times \bar{P}\) So both \(\bar{L}\) and \(\vec{\tau}\) depend on axis about which they are measured.
366179
In an orbital motion, the angular momentum vector is:
1 Parallel to the linear momentum
2 Along the radius vector
3 Perpendicular to the orbital plane
4 In the orbital plane
Explanation:
Option (3) is correct
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
366180
Assertion : A particle moving on a straight line with a uniform velocity, its angular momentum is always zero. Reason : The angular momentum is a vector.
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:
\(|\vec{L}|=m v r \sin \theta\) If straight line, \(\theta=0\) \(\Rightarrow L=0\). Angular momentum is zero only if the particle's velocity vector is parallel to the position vector so assertion is wrong. The angular momentum is a vector quantity. So correct option is (4).
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
366181
A particle moves in a straight line with constant acceleration. Its angular momentum about any fixed point is :
1 constant
2 directly proportional to time
3 inversely proportional to time
4 inversely proportional to distance covered
Explanation:
Let a particle \({A}\) of mass \({m}\) is moving in a straight line with velocity \({v}\) as shown below. Angular momentum of particle about any fixed point say \({O}\) will be \(\begin{aligned}& L=m\left(\overrightarrow{O A}^{\prime} \times \vec{v}\right) \\& L=m v O A^{\prime} \sin \theta=m v r \\& \Rightarrow L=m(a t) r \quad[a=\text { constant acceleration }] \\& L=(m a r) t \\& \therefore L \propto t\end{aligned}\). So correct option is (2)
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
366182
Statement A : Angular momentum of a body depends on axis of measurement. Statement B : Torque acting on a body depends on axis of measurement
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both statements are correct.
4 Both Statements are incorrect.
Explanation:
\(\bar{L}=\bar{r} \times \bar{P}, \vec{\tau}=\bar{r} \times \bar{P}\) So both \(\bar{L}\) and \(\vec{\tau}\) depend on axis about which they are measured.
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
366179
In an orbital motion, the angular momentum vector is:
1 Parallel to the linear momentum
2 Along the radius vector
3 Perpendicular to the orbital plane
4 In the orbital plane
Explanation:
Option (3) is correct
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
366180
Assertion : A particle moving on a straight line with a uniform velocity, its angular momentum is always zero. Reason : The angular momentum is a vector.
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:
\(|\vec{L}|=m v r \sin \theta\) If straight line, \(\theta=0\) \(\Rightarrow L=0\). Angular momentum is zero only if the particle's velocity vector is parallel to the position vector so assertion is wrong. The angular momentum is a vector quantity. So correct option is (4).
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
366181
A particle moves in a straight line with constant acceleration. Its angular momentum about any fixed point is :
1 constant
2 directly proportional to time
3 inversely proportional to time
4 inversely proportional to distance covered
Explanation:
Let a particle \({A}\) of mass \({m}\) is moving in a straight line with velocity \({v}\) as shown below. Angular momentum of particle about any fixed point say \({O}\) will be \(\begin{aligned}& L=m\left(\overrightarrow{O A}^{\prime} \times \vec{v}\right) \\& L=m v O A^{\prime} \sin \theta=m v r \\& \Rightarrow L=m(a t) r \quad[a=\text { constant acceleration }] \\& L=(m a r) t \\& \therefore L \propto t\end{aligned}\). So correct option is (2)
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
366182
Statement A : Angular momentum of a body depends on axis of measurement. Statement B : Torque acting on a body depends on axis of measurement
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both statements are correct.
4 Both Statements are incorrect.
Explanation:
\(\bar{L}=\bar{r} \times \bar{P}, \vec{\tau}=\bar{r} \times \bar{P}\) So both \(\bar{L}\) and \(\vec{\tau}\) depend on axis about which they are measured.
