Q.32. The pulley shown in figure is made using a thin rim and two rods of length equal to diameter o

Q: What is the correct answer for JEE Main Physics Q32?

The correct answer is C. Let the radius of the rim be R. The diameter is 2R.Each rod has length 2R.Mass of rim = M.Mass of each rod = M.Total mass of pulley $M_P = M_{rim} + 2 M_{rod} = M + 2M = 3M$.Moment of inertia of the rim (ring) about its center is $I_{rim} = MR^2$.Each rod passes through the center. For a rod of length L and mass M, its moment of inertia about its center perpendicular to its length is $\frac{ML^2}{

Q: Q.32. The pulley shown in figure is made using a thin rim and two rods of length equal to diameter of the rim. The rim and each rod have a mass of M. Two blocks of mass of M and m are attached to two…

The correct answer is C. Let the radius of the rim be R. The diameter is 2R.Each rod has length 2R.Mass of rim = M.Mass of each rod = M.Total mass of pulley $M_P = M_{rim} + 2 M_{rod} = M + 2M = 3M$.Moment of inertia of the rim (ring) about its center is $I_{rim} = MR^2$.Each rod passes through the center. For a rod of length L and mass M, its moment of inertia about its center perpendicular to its length is $\frac{ML^2}{

Q: What are the options for this JEE Main Physics question?

A. $$\frac{(M-m)g}{\frac{13}{6}M+m}$$ · B. $$\frac{(M-m)g}{M+m}$$ · C. $$\frac{(M-m)g}{\frac{8}{3}M+m}$$ · D. $$\frac{(M-m)g}{2M+m}$$

Question

Q.32. The pulley shown in figure is made using a thin rim and two rods of length equal to diameter of the rim. The rim and each rod have a mass of M. Two blocks of mass of M and m are attached to two ends of a light string passing over the pulley, which is hinged to rotate freely in vertical plane about its centre. The magnitudes of the acceleration experienced by the blocks is .......... (assume no slipping of string on pulley.) [Figure: Pulley system]

FirstInTest · Accepted Answer

Let the radius of the rim be R. The diameter is 2R.Each rod has length 2R.Mass of rim = M.Mass of each rod = M.Total mass of pulley $M_P = M_{rim} + 2 M_{rod} = M + 2M = 3M$.Moment of inertia of the rim (ring) about its center is $I_{rim} = MR^2$.Each rod passes through the center. For a rod of length L and mass M, its moment of inertia about its center perpendicular to its length is $\frac{ML^2}{12}$. Here $L=2R$.So, for one rod, $I_{rod} = \frac{M(2R)^2}{12} = \frac{4MR^2}{12} = \frac{MR^2}{3}$.For two rods, $I_{2rods} = 2 	imes \frac{MR^2}{3} = \frac{2MR^2}{3}$.Total moment of inertia of the pulley $I_{pulley} = I_{rim} + I_{2rods} = MR^2 + \frac{2MR^2}{3} = \frac{5MR^2}{3}$.Let $T_1$ be the tension on the side of mass M and $T_2$ be the tension on the side of mass m.Let the acceleration of the blocks be a. For mass M: $Mg - T_1 = Ma$ (assuming M > m, so M moves down)For mass m: $T_2 - mg = ma$For the pulley: $(T_1 - T_2)R = I_{pulley} \alpha$.Since there is no slipping, $a = R\alpha$, so $\alpha = a/R$.$(T_1 - T_2)R = \frac{5MR^2}{3} \frac{a}{R}$.$T_1 - T_2 = \frac{5Ma}{3}$.Add the three equations:$(Mg - T_1) + (T_2 - mg) + (T_1 - T_2) = Ma + ma + \frac{5Ma}{3}$.$Mg - mg = (M+m)a + \frac{5Ma}{3}$.$(M-m)g = a (M+m + \frac{5M}{3})$.$(M-m)g = a (\frac{3M+3m+5M}{3})$.$(M-m)g = a (\frac{8M+3m}{3})$.$a = \frac{3(M-m)g}{8M+3m}$.The provided solution in the image has a slightly different approach for the moment of inertia and the final expression.Let's re-check the solution's steps:The solution calculates $I = Mr^2 + \frac{M(2r)^2}{12} 	imes 2$. This is $I = MR^2 + \frac{8MR^2}{12} = MR^2 + \frac{2MR^2}{3} = \frac{5MR^2}{3}$. This matches my calculation.Then it uses $(M-m)g = (M+m + \frac{I}{R^2})a$.This is derived from $T_1 - T_2 = \frac{I}{R^2}a$.So $(M-m)g = (M+m)a + \frac{I}{R^2}a$.Substituting $I = \frac{5MR^2}{3}$:$(M-m)g = (M+m)a + \frac{5MR^2}{3R^2}a = (M+m)a + \frac{5M}{3}a$.$(M-m)g = a(M+m+\frac{5M}{3})$.$(M-m)g = a(\frac{3M+3m+5M}{3}) = a(\frac{8M+3m}{3})$.$a = \frac{3(M-m)g}{8M+3m}$.This result is not among the options.Let's re-examine the options and the provided answer (3).Option (3) is $\frac{8(M-m)g}{M+m}$. This is incorrect.Option (3) in the image is $\frac{(M-m)g}{\frac{8}{3}M+m}$. This is equivalent to $\frac{3(M-m)g}{8M+3m}$.So, my calculation matches option (3). The OCR for option (3) was a bit off. The correct option (3) is $\frac{(M-m)g}{\frac{8}{3}M+m}$.

Answer

$$\frac{(M-m)g}{\frac{13}{6}M+m}$$

Answer

$$\frac{(M-m)g}{M+m}$$

Answer

$$\frac{(M-m)g}{\frac{8}{3}M+m}$$

Answer

$$\frac{(M-m)g}{2M+m}$$

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