Question

JEE MAINS PHYSICS QUESTION SEPTEMBER 2020

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{\frac{I\:F\:v^{2}}{E\:L^{4}}=Young\:modulus}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies I= Moment \: of \: inertia \\ \\ \tt: \implies F= Force \\ \\ \tt: \implies v = Velocity \\ \\ \tt: \implies E = Energy \\ \\ \tt: \implies L = Length \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Dimension \: of \: \: \: \frac{I\: F\: {v}^{2} }{E \: {L}^{4} } =?$

• According to given question :

$\tt \circ \:Dimension \: of \: moment \: of \: inertia = M {L}^{2} {T}^{0} \\ \\ \tt \circ \:Dimension \: of \: force = M {L} {T}^{ - 2 } \\ \\\tt \circ \:Dimension \: of \: velocity = {M}^{0} {L} {T}^{ - 1} \\ \\\tt \circ \:Dimension \: of \: energy = M {L}^{2} {T}^{ - 2} \\ \\ \tt \circ \:Dimension \: of \: length = {M}^{0} {L} {T}^{0} \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies \frac{I \: F\: {v}^{2} }{E \: {L}^{4} } = \frac{M {L}^{2} {T}^{0} \times {MLT}^{ - 2} \times ( {M}^{0} L{T} ^{ - 1})^{2} }{ {ML}^{2} {T}^{ - 2} \times ({M}^{0} {L} {T}^{0} )^{4} } \\ \\ \tt: \implies \frac{I\: F\: {v}^{2} }{E \: {L}^{4} } = \frac{M {L}^{2} \times {MLT}^{ - 2} \times {L}^{2} {T}^{ - 2} }{M {L}^{2} {T}^{ - 2} \times {L}^{4} } \\ \\ \tt: \implies \frac{I \: F \: {v}^{2} }{E \: {L}^{4} } = \frac{M}{ {T}^{2} } \\ \\ \green{\tt: \implies \frac{I\: F \: {v}^{2} }{E \: {L}^{4} } = {MT}^{ - 2} = Young \: modulus}$

Answer 2

Option (3) Young modulus.

✔ Explanation :-

✔ Given :-

• I is a moment of inertia.
• F is force.
• v is velocity.
• E is energy.
• L is length.

✔ To Find :-

• $\tt \: Dimension \: \: of \: \: \frac{IFv {}^{2} }{EL{}^{4} }$

✔ Solution :-

We know that,

• $\tt \: \: Dimesion \: \: of \: \: moment \: \: of \: \: inertia \: = \: ML {}^{2} T {}^{0}$
• $\tt \: \: Dimesion \: \: of \: \: force \: = \: MLT {}^{ - 2}$
• $\tt \: \: Dimesion \: \: of \: \: \: velocity \: = \: M {}^{0} LT {}^{ - 1}$
• $\tt \: \: Dimesion \: \: of \: \: energy\: = \: ML {}^{2} T {}^{ - 2}$
• $\tt \: \: Dimesion \: \: of \: \:length\: = \: M {}^{0} LT {}^{ 0}$

Now,

$\tt \frac{IFv {}^{2} }{EL{}^{4} } = \frac{ML {}^{2} T {}^{0} \times \: MLT {}^{ - 2} \times \: (M {}^{0} LT {}^{ - 1} ) {}^{2} }{ML {}^{2} T {}^{ - 2} \times M {}^{0} LT {}^{ 0} }$

$\tt \frac{IFv {}^{2} }{EL{}^{4} } = \frac{ML {}^{2} \times \: MLT {}^{ - 2} \times \: L {}^{2} T {}^{ -2} }{ML {}^{2} T {}^{ - 2} \times{ L {}^{4} }}$

$\tt \frac{IFv {}^{2} }{EL{}^{4} } = \frac{M}{T {}^{2} }$

Hence,

${ \boxed{ \red{\tt \frac{IFv {}^{2} }{EL{}^{4} } =MT {}^{ - 2} = young \: \: modulus}}}$