JEE Physics Units And Measurements MCQs Set 04

Answer: (a) \( \frac{\pi}{90} \)

Explanation:
1. \( \sin 2^\circ\cos 2^\circ = \tfrac12 \sin 4^\circ \)
2. For a small angle, \( \sin\theta \approx \theta \) in radians, and \( 4^\circ = \frac{4\pi}{180} = \frac{\pi}{45} \) rad
3. So \( x \approx \frac12 \cdot \frac{\pi}{45} = \frac{\pi}{90} \). Answer: (a)

Teacher's Note:
1. Use the double-angle identity first, then the small-angle approximation.
2. The angle must be in radians before replacing sin with the angle.

Answer: (c) \( M^{-1}L^0T^0 \)

Explanation:
1. \( \int \vec F\cdot d\vec s \) is work, so the left side is power \( = [ML^2T^{-3}] \)
2. \( [FP] = [MLT^{-2}][MLT^{-1}] = [M^2L^2T^{-3}] \)
3. \( [A] = \frac{ML^2T^{-3}}{M^2L^2T^{-3}} = M^{-1} \). Answer: (c)

Teacher's Note:
1. Identify the physical quantity on each side before comparing dimensions.
2. Rate of change of work is power.

Answer: (a) \( ML^3T^{-3}I^{-1} \)

Explanation:
1. Electric flux \( = E \times \text{area} \)
2. \( [E] = \frac{[F]}{[q]} = \frac{MLT^{-2}}{IT} = MLT^{-3}I^{-1} \)
3. Multiply by \( L^2 \): \( ML^3T^{-3}I^{-1} \). Answer: (a)

Teacher's Note:
1. Start from the dimensions of electric field.
2. Charge has dimensions of current times time.

Answer: (b) \( 2\pi x = \sqrt{z/y} \)

Explanation:
1. The table gives \( [x] = [T^{-1}] \), \( [y] = [L] \) and \( [z] = [LT^{-2}] \)
2. \( \frac{z}{y} = T^{-2} \), so \( \sqrt{z/y} = T^{-1} \), which equals \( [x] \)
3. The other options give wrong dimensions, so only \( 2\pi x = \sqrt{z/y} \) is dimensionally consistent. Answer: (b)

Teacher's Note:
1. Check each option's dimensions on both sides.
2. A pure number such as 2 pi has no dimensions.

Answer: (d) \( L^{-2}T^2 \)

Explanation:
1. \( \alpha = \frac{\text{density}}{P} \)
2. \( [\text{density}] = ML^{-3} \) and \( [P] = ML^{-1}T^{-2} \)
3. \( [\alpha] = \frac{ML^{-3}}{ML^{-1}T^{-2}} = L^{-2}T^2 \). Answer: (d)

Teacher's Note:
1. Rearrange for the unknown quantity before substituting dimensions.
2. Pressure is force per unit area.

Answer: (b) \( \left[M^{-1/2}\,T^{1}\,W^{1/2}\right] \)

Explanation:
1. \( W = ML^2T^{-2} \)
2. So \( L^2 = \frac{W T^2}{M} \)
3. \( L = M^{-1/2}\,T\,W^{1/2} \). Answer: (b)

Teacher's Note:
1. Express work in terms of M, L and T, then solve for L.
2. Take the square root for the final dimensional formula.

Answer: (c) \( L^2M^0T^0 \)

Explanation:
1. The right side must be dimensionless, so \( [\alpha z] = [k_B\theta] = \) energy \( = ML^2T^{-2} \)
2. \( [\alpha] = \frac{ML^2T^{-2}}{L} = MLT^{-2} \)
3. \( \frac{\alpha}{p\beta} \) is dimensionless, so \( [\beta] = \frac{[\alpha]}{[p]} = \frac{MLT^{-2}}{ML^{-1}T^{-2}} = L^2 \). Answer: (c)

Teacher's Note:
1. The argument of a logarithm is always dimensionless.
2. Use the second exponent to get alpha, then the first to get beta.