363467
If a street light of mass \(M\) is suspended from the end of a uniform rod of length \(L\) in different possible patterns as shown in figure, then
1 Pattern \(A\) is more sturdy
2 Pattern \(B\) is more sturdy
3 Pattern \(C\) is more sturdy
4 All will have same sturdiness
Explanation:
For equilibrium of street light, net torque on rod should be zero. \(m g \times x=T \times y\) or \(T=\dfrac{m g x}{y}\) For \(T\) to be minimum, \(y\) should be maximum. Hence, pattern \(A\) is more study.
AIIMS - 2006
PHXI05:LAWS OF MOTION
363468
A body of mass \(500\;g\) moves along \(x\)-axis such that it's velocity varies with displacement \(x\) according to the relation \(v = 10\sqrt x \;m{\rm{/}}s\), the force acting on the body is
1 \(5\;N\)
2 \(125\;N\)
3 \(166\;N\)
4 \(25\;N\)
Explanation:
Mass, \(m = 500\;g = 0.5\;kg\) velocity \(v = 10\sqrt x \;m{\rm{/}}s\) \(a=\dfrac{d \vec{v}}{d t}=\dfrac{d \vec{v}}{d x} \cdot\left(\dfrac{d x}{d t}\right)=\dfrac{1}{2}(10) x^{-1 / 2} \dfrac{d x}{d t}\) \( = \frac{1}{2}\frac{{(10)}}{{\sqrt x }} \cdot v = \frac{1}{2}\frac{{(10)}}{{\sqrt x }}10(\sqrt x )\,\,;\) \(a = 50\;m{\rm{/}}{s^2}\) Force, \(F=m a=0.5 \times 50=25 N\) \(\therefore \vec F = 25\,\hat i\,N.\)
JEE - 2023
PHXI05:LAWS OF MOTION
363469
Assertion : A man in a closed cabin falling freely does not experience gravity. Reason : Inertia and gravitational mass have equivalence.
1 Both assertion and reason are correct and reason is the correct explanation of assertion.
2 Both assertion and reason are correct but reason is not the correct explanation of assertion.
3 Assertion is correct but reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
\({m_{grav.}}g - N = {m_{inertial}}\,.\,\,a\) For freely falling \(a = g\). Since \({m_{grav}} = {m_{inert}} \Rightarrow N = 0\)
PHXI05:LAWS OF MOTION
363470
Two forces having magnitude \(A\) and \(\dfrac{A}{2}\) are perpendicular to each other. The magnitude of their resultant is
1 \(\dfrac{5 A}{2}\)
2 \(\dfrac{\sqrt{5} A^{2}}{2}\)
3 \(\dfrac{\sqrt{5} A}{4}\)
4 \(\dfrac{\sqrt{5} A}{2}\)
Explanation:
Force is vector quantity where resultant of two forces is given as By parallelogram Law, \(F^{2}=F_{1}^{2}+F_{2}^{2}+2 F_{1} F_{2} \cos \theta\) \(=(A)^{2}+\left(\dfrac{A}{2}\right)^{2}+2 \times A \times \dfrac{A}{2} \times \cos 90^{\circ}\) (As forces are acting perpendicularly) \(F^{2}=\dfrac{5 A^{2}}{4} \Rightarrow F=\dfrac{\sqrt{5} A}{2}\) units
JEE - 2023
PHXI05:LAWS OF MOTION
363471
Assertion : From Newton's second law of motion impulse is equal to change in momentum. Reason : Impulse and momentum have different dimensions.
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:
Impulse \(=F(\Delta t)=\Delta p\) Since impulse equals the change in momentum, impulse and momentum have the same dimensions. So correct option is (3)
363467
If a street light of mass \(M\) is suspended from the end of a uniform rod of length \(L\) in different possible patterns as shown in figure, then
1 Pattern \(A\) is more sturdy
2 Pattern \(B\) is more sturdy
3 Pattern \(C\) is more sturdy
4 All will have same sturdiness
Explanation:
For equilibrium of street light, net torque on rod should be zero. \(m g \times x=T \times y\) or \(T=\dfrac{m g x}{y}\) For \(T\) to be minimum, \(y\) should be maximum. Hence, pattern \(A\) is more study.
AIIMS - 2006
PHXI05:LAWS OF MOTION
363468
A body of mass \(500\;g\) moves along \(x\)-axis such that it's velocity varies with displacement \(x\) according to the relation \(v = 10\sqrt x \;m{\rm{/}}s\), the force acting on the body is
1 \(5\;N\)
2 \(125\;N\)
3 \(166\;N\)
4 \(25\;N\)
Explanation:
Mass, \(m = 500\;g = 0.5\;kg\) velocity \(v = 10\sqrt x \;m{\rm{/}}s\) \(a=\dfrac{d \vec{v}}{d t}=\dfrac{d \vec{v}}{d x} \cdot\left(\dfrac{d x}{d t}\right)=\dfrac{1}{2}(10) x^{-1 / 2} \dfrac{d x}{d t}\) \( = \frac{1}{2}\frac{{(10)}}{{\sqrt x }} \cdot v = \frac{1}{2}\frac{{(10)}}{{\sqrt x }}10(\sqrt x )\,\,;\) \(a = 50\;m{\rm{/}}{s^2}\) Force, \(F=m a=0.5 \times 50=25 N\) \(\therefore \vec F = 25\,\hat i\,N.\)
JEE - 2023
PHXI05:LAWS OF MOTION
363469
Assertion : A man in a closed cabin falling freely does not experience gravity. Reason : Inertia and gravitational mass have equivalence.
1 Both assertion and reason are correct and reason is the correct explanation of assertion.
2 Both assertion and reason are correct but reason is not the correct explanation of assertion.
3 Assertion is correct but reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
\({m_{grav.}}g - N = {m_{inertial}}\,.\,\,a\) For freely falling \(a = g\). Since \({m_{grav}} = {m_{inert}} \Rightarrow N = 0\)
PHXI05:LAWS OF MOTION
363470
Two forces having magnitude \(A\) and \(\dfrac{A}{2}\) are perpendicular to each other. The magnitude of their resultant is
1 \(\dfrac{5 A}{2}\)
2 \(\dfrac{\sqrt{5} A^{2}}{2}\)
3 \(\dfrac{\sqrt{5} A}{4}\)
4 \(\dfrac{\sqrt{5} A}{2}\)
Explanation:
Force is vector quantity where resultant of two forces is given as By parallelogram Law, \(F^{2}=F_{1}^{2}+F_{2}^{2}+2 F_{1} F_{2} \cos \theta\) \(=(A)^{2}+\left(\dfrac{A}{2}\right)^{2}+2 \times A \times \dfrac{A}{2} \times \cos 90^{\circ}\) (As forces are acting perpendicularly) \(F^{2}=\dfrac{5 A^{2}}{4} \Rightarrow F=\dfrac{\sqrt{5} A}{2}\) units
JEE - 2023
PHXI05:LAWS OF MOTION
363471
Assertion : From Newton's second law of motion impulse is equal to change in momentum. Reason : Impulse and momentum have different dimensions.
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:
Impulse \(=F(\Delta t)=\Delta p\) Since impulse equals the change in momentum, impulse and momentum have the same dimensions. So correct option is (3)
363467
If a street light of mass \(M\) is suspended from the end of a uniform rod of length \(L\) in different possible patterns as shown in figure, then
1 Pattern \(A\) is more sturdy
2 Pattern \(B\) is more sturdy
3 Pattern \(C\) is more sturdy
4 All will have same sturdiness
Explanation:
For equilibrium of street light, net torque on rod should be zero. \(m g \times x=T \times y\) or \(T=\dfrac{m g x}{y}\) For \(T\) to be minimum, \(y\) should be maximum. Hence, pattern \(A\) is more study.
AIIMS - 2006
PHXI05:LAWS OF MOTION
363468
A body of mass \(500\;g\) moves along \(x\)-axis such that it's velocity varies with displacement \(x\) according to the relation \(v = 10\sqrt x \;m{\rm{/}}s\), the force acting on the body is
1 \(5\;N\)
2 \(125\;N\)
3 \(166\;N\)
4 \(25\;N\)
Explanation:
Mass, \(m = 500\;g = 0.5\;kg\) velocity \(v = 10\sqrt x \;m{\rm{/}}s\) \(a=\dfrac{d \vec{v}}{d t}=\dfrac{d \vec{v}}{d x} \cdot\left(\dfrac{d x}{d t}\right)=\dfrac{1}{2}(10) x^{-1 / 2} \dfrac{d x}{d t}\) \( = \frac{1}{2}\frac{{(10)}}{{\sqrt x }} \cdot v = \frac{1}{2}\frac{{(10)}}{{\sqrt x }}10(\sqrt x )\,\,;\) \(a = 50\;m{\rm{/}}{s^2}\) Force, \(F=m a=0.5 \times 50=25 N\) \(\therefore \vec F = 25\,\hat i\,N.\)
JEE - 2023
PHXI05:LAWS OF MOTION
363469
Assertion : A man in a closed cabin falling freely does not experience gravity. Reason : Inertia and gravitational mass have equivalence.
1 Both assertion and reason are correct and reason is the correct explanation of assertion.
2 Both assertion and reason are correct but reason is not the correct explanation of assertion.
3 Assertion is correct but reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
\({m_{grav.}}g - N = {m_{inertial}}\,.\,\,a\) For freely falling \(a = g\). Since \({m_{grav}} = {m_{inert}} \Rightarrow N = 0\)
PHXI05:LAWS OF MOTION
363470
Two forces having magnitude \(A\) and \(\dfrac{A}{2}\) are perpendicular to each other. The magnitude of their resultant is
1 \(\dfrac{5 A}{2}\)
2 \(\dfrac{\sqrt{5} A^{2}}{2}\)
3 \(\dfrac{\sqrt{5} A}{4}\)
4 \(\dfrac{\sqrt{5} A}{2}\)
Explanation:
Force is vector quantity where resultant of two forces is given as By parallelogram Law, \(F^{2}=F_{1}^{2}+F_{2}^{2}+2 F_{1} F_{2} \cos \theta\) \(=(A)^{2}+\left(\dfrac{A}{2}\right)^{2}+2 \times A \times \dfrac{A}{2} \times \cos 90^{\circ}\) (As forces are acting perpendicularly) \(F^{2}=\dfrac{5 A^{2}}{4} \Rightarrow F=\dfrac{\sqrt{5} A}{2}\) units
JEE - 2023
PHXI05:LAWS OF MOTION
363471
Assertion : From Newton's second law of motion impulse is equal to change in momentum. Reason : Impulse and momentum have different dimensions.
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:
Impulse \(=F(\Delta t)=\Delta p\) Since impulse equals the change in momentum, impulse and momentum have the same dimensions. So correct option is (3)
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI05:LAWS OF MOTION
363467
If a street light of mass \(M\) is suspended from the end of a uniform rod of length \(L\) in different possible patterns as shown in figure, then
1 Pattern \(A\) is more sturdy
2 Pattern \(B\) is more sturdy
3 Pattern \(C\) is more sturdy
4 All will have same sturdiness
Explanation:
For equilibrium of street light, net torque on rod should be zero. \(m g \times x=T \times y\) or \(T=\dfrac{m g x}{y}\) For \(T\) to be minimum, \(y\) should be maximum. Hence, pattern \(A\) is more study.
AIIMS - 2006
PHXI05:LAWS OF MOTION
363468
A body of mass \(500\;g\) moves along \(x\)-axis such that it's velocity varies with displacement \(x\) according to the relation \(v = 10\sqrt x \;m{\rm{/}}s\), the force acting on the body is
1 \(5\;N\)
2 \(125\;N\)
3 \(166\;N\)
4 \(25\;N\)
Explanation:
Mass, \(m = 500\;g = 0.5\;kg\) velocity \(v = 10\sqrt x \;m{\rm{/}}s\) \(a=\dfrac{d \vec{v}}{d t}=\dfrac{d \vec{v}}{d x} \cdot\left(\dfrac{d x}{d t}\right)=\dfrac{1}{2}(10) x^{-1 / 2} \dfrac{d x}{d t}\) \( = \frac{1}{2}\frac{{(10)}}{{\sqrt x }} \cdot v = \frac{1}{2}\frac{{(10)}}{{\sqrt x }}10(\sqrt x )\,\,;\) \(a = 50\;m{\rm{/}}{s^2}\) Force, \(F=m a=0.5 \times 50=25 N\) \(\therefore \vec F = 25\,\hat i\,N.\)
JEE - 2023
PHXI05:LAWS OF MOTION
363469
Assertion : A man in a closed cabin falling freely does not experience gravity. Reason : Inertia and gravitational mass have equivalence.
1 Both assertion and reason are correct and reason is the correct explanation of assertion.
2 Both assertion and reason are correct but reason is not the correct explanation of assertion.
3 Assertion is correct but reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
\({m_{grav.}}g - N = {m_{inertial}}\,.\,\,a\) For freely falling \(a = g\). Since \({m_{grav}} = {m_{inert}} \Rightarrow N = 0\)
PHXI05:LAWS OF MOTION
363470
Two forces having magnitude \(A\) and \(\dfrac{A}{2}\) are perpendicular to each other. The magnitude of their resultant is
1 \(\dfrac{5 A}{2}\)
2 \(\dfrac{\sqrt{5} A^{2}}{2}\)
3 \(\dfrac{\sqrt{5} A}{4}\)
4 \(\dfrac{\sqrt{5} A}{2}\)
Explanation:
Force is vector quantity where resultant of two forces is given as By parallelogram Law, \(F^{2}=F_{1}^{2}+F_{2}^{2}+2 F_{1} F_{2} \cos \theta\) \(=(A)^{2}+\left(\dfrac{A}{2}\right)^{2}+2 \times A \times \dfrac{A}{2} \times \cos 90^{\circ}\) (As forces are acting perpendicularly) \(F^{2}=\dfrac{5 A^{2}}{4} \Rightarrow F=\dfrac{\sqrt{5} A}{2}\) units
JEE - 2023
PHXI05:LAWS OF MOTION
363471
Assertion : From Newton's second law of motion impulse is equal to change in momentum. Reason : Impulse and momentum have different dimensions.
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:
Impulse \(=F(\Delta t)=\Delta p\) Since impulse equals the change in momentum, impulse and momentum have the same dimensions. So correct option is (3)
363467
If a street light of mass \(M\) is suspended from the end of a uniform rod of length \(L\) in different possible patterns as shown in figure, then
1 Pattern \(A\) is more sturdy
2 Pattern \(B\) is more sturdy
3 Pattern \(C\) is more sturdy
4 All will have same sturdiness
Explanation:
For equilibrium of street light, net torque on rod should be zero. \(m g \times x=T \times y\) or \(T=\dfrac{m g x}{y}\) For \(T\) to be minimum, \(y\) should be maximum. Hence, pattern \(A\) is more study.
AIIMS - 2006
PHXI05:LAWS OF MOTION
363468
A body of mass \(500\;g\) moves along \(x\)-axis such that it's velocity varies with displacement \(x\) according to the relation \(v = 10\sqrt x \;m{\rm{/}}s\), the force acting on the body is
1 \(5\;N\)
2 \(125\;N\)
3 \(166\;N\)
4 \(25\;N\)
Explanation:
Mass, \(m = 500\;g = 0.5\;kg\) velocity \(v = 10\sqrt x \;m{\rm{/}}s\) \(a=\dfrac{d \vec{v}}{d t}=\dfrac{d \vec{v}}{d x} \cdot\left(\dfrac{d x}{d t}\right)=\dfrac{1}{2}(10) x^{-1 / 2} \dfrac{d x}{d t}\) \( = \frac{1}{2}\frac{{(10)}}{{\sqrt x }} \cdot v = \frac{1}{2}\frac{{(10)}}{{\sqrt x }}10(\sqrt x )\,\,;\) \(a = 50\;m{\rm{/}}{s^2}\) Force, \(F=m a=0.5 \times 50=25 N\) \(\therefore \vec F = 25\,\hat i\,N.\)
JEE - 2023
PHXI05:LAWS OF MOTION
363469
Assertion : A man in a closed cabin falling freely does not experience gravity. Reason : Inertia and gravitational mass have equivalence.
1 Both assertion and reason are correct and reason is the correct explanation of assertion.
2 Both assertion and reason are correct but reason is not the correct explanation of assertion.
3 Assertion is correct but reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
\({m_{grav.}}g - N = {m_{inertial}}\,.\,\,a\) For freely falling \(a = g\). Since \({m_{grav}} = {m_{inert}} \Rightarrow N = 0\)
PHXI05:LAWS OF MOTION
363470
Two forces having magnitude \(A\) and \(\dfrac{A}{2}\) are perpendicular to each other. The magnitude of their resultant is
1 \(\dfrac{5 A}{2}\)
2 \(\dfrac{\sqrt{5} A^{2}}{2}\)
3 \(\dfrac{\sqrt{5} A}{4}\)
4 \(\dfrac{\sqrt{5} A}{2}\)
Explanation:
Force is vector quantity where resultant of two forces is given as By parallelogram Law, \(F^{2}=F_{1}^{2}+F_{2}^{2}+2 F_{1} F_{2} \cos \theta\) \(=(A)^{2}+\left(\dfrac{A}{2}\right)^{2}+2 \times A \times \dfrac{A}{2} \times \cos 90^{\circ}\) (As forces are acting perpendicularly) \(F^{2}=\dfrac{5 A^{2}}{4} \Rightarrow F=\dfrac{\sqrt{5} A}{2}\) units
JEE - 2023
PHXI05:LAWS OF MOTION
363471
Assertion : From Newton's second law of motion impulse is equal to change in momentum. Reason : Impulse and momentum have different dimensions.
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:
Impulse \(=F(\Delta t)=\Delta p\) Since impulse equals the change in momentum, impulse and momentum have the same dimensions. So correct option is (3)
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