Young's Modulus Torque Viscosity and G Dimensions JEE Main

 

❓ Question

Match the following:

List I	List II
(A) Young's Modulus $(Y)$	(1) $[ML^{-1}T^{-1}]$
(B) Torque	(2) $[ML^{-1}T^{-2}]$
(C) Coefficient of Viscosity	(3) $[M^{-1}L^3T^{-2}]$
(D) Gravitational Constant	(4) $[ML^2T^{-2}]$

---

✍️ Solution

(A) Young's Modulus $(Y)$

Young's modulus is

$$Y=\frac{\text{Stress}}{\text{Strain}}$$

Since strain is dimensionless,

$$[Y]=[\text{Stress}] =\frac{[F]}{[A]}$$
$$[Y]=\frac{[MLT^{-2}]}{[L^2]}$$
$$\boxed{[Y]=[ML^{-1}T^{-2}]}$$

Hence,

$$\boxed{A\rightarrow2}$$

---

(B) Torque

Torque is

$$\tau=F\times r$$

Therefore,

$$[\tau]=[MLT^{-2}][L]$$
$$\boxed{[\tau]=[ML^2T^{-2}]}$$

Hence,

$$\boxed{B\rightarrow4}$$

---

(C) Coefficient of Viscosity $(\eta)$

Using the viscous force relation,

$$F=\eta A\frac{dv}{dx}$$

Therefore,

$$\eta=\frac{F}{A\left(\frac{dv}{dx}\right)}$$

Now,

$$\left[\frac{dv}{dx}\right]=[T^{-1}]$$

Thus,

$$[\eta] = \frac{[MLT^{-2}]}{[L^2][T^{-1}]}$$
$$\boxed{[\eta]=[ML^{-1}T^{-1}]}$$

Hence,

$$\boxed{C\rightarrow1}$$

---

(D) Gravitational Constant $(G)$

From Newton's law of gravitation,

$$F=\frac{Gm_1m_2}{r^2}$$

Therefore,

$$G=\frac{Fr^2}{m_1m_2}$$
$$[G] = \frac{[MLT^{-2}][L^2]}{[M][M]}$$
$$\boxed{[G]=[M^{-1}L^3T^{-2}]}$$

Hence,

$$\boxed{D\rightarrow3}$$

---

✅ Final Answer

$$\boxed{A\rightarrow2,\quad B\rightarrow4,\quad C\rightarrow1,\quad D\rightarrow3}$$