JEE Main · 2024 · 08 Apr 2024 · Shift 1 · Units & Measurement

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The diameter of a sphere is measured using a vernier caliper whose 9 divisions of main scale are equal to 10 divisions of vernier scale. The shortest division on the main scale is equal to 1 mm. The main scale reading is 2 cm and second division of vernier scale coincides with a division on main scale. If mass of the sphere is 8.635 g, the density of the sphere is:

(A) $2.5 g / cm^{3}$
(B) $1.7 g / cm^{3}$
(C) $2.2 g / cm^{3}$
(D) $2.0 g / cm^{3}$ ✓ Correct

Answer: (D)

Solution

Least count

= 1 MSD - 1 VSD = (1 - \frac{9}{10}) MSD = \frac{1}{10} (1 mm) = 0.01 cm

.
Diameter

= MSR + LC \times VSR = 2 + 0.01 (2) = 2.02 cm

V = \frac{4}{3} π (\frac{2.02}{2})^{3} \approx 4.32 cm^{3}

, so

ρ = \frac{8.635}{4.32} \approx 2.0 g/cm^{3}

. Option (D).

JEE Main · 2024 · 08 Apr 2024 · Shift 1 · Single correct option

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