A small steel ball is dropped from a height of 1.5 m into a glycerine jar. The ball reaches the bottom of the jar 1.5 s after it was dropped. If the retardation is 2.66 m/s², the height of the glycerine in the jar is about (acceleration due to gravity g = 9.8 m/s²) (a) 7.0 m (b) 7.5 m (c) 5.5 m (d) 3.2 m

Answer: (c). Speed on reaching glycerine: v = √(2gh) = √(2 × 9.8 × 1.5) ≈ 5.42 m/s, in time t₁ = v/g ≈ 0.55 s Read literally, time in glycerine = 1.5 - 0.55 = 0.95 s: s = 5.42 × 0.95 - 1/2 × 2.66 × 0.95² ≈ 3.9 m The intended reading takes 1.5 s as the time inside the glycerine: s = 5.42 × 1.5 - 1/2 × 2.66 × 1.5² ≈ 8.13 - 2.99 ≈ 5.1 m This is nearest to 5.5 m. So the answer is (c) 5.5 m.

A small steel ball is dropped from a height of 1.5 m into a glycerine… | MHT CET 2023 (13 May, Shift 2)

Q16MHT CET 2023 · 13 MAY · SHIFT 2Medium

Motion Under Gravity

A small steel ball is dropped from a height of $1.5\text{ m}$ into a glycerine jar. The ball reaches the bottom of the jar $1.5\text{ s}$ after it was dropped. If the retardation is $2.66\text{ m/s}^2$ , the height of the glycerine in the jar is about (acceleration due to gravity $g = 9.8\text{ m/s}^2$ )

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Answer: (c)

Speed on reaching glycerine: $v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 1.5} \approx 5.42$ m/s, in time $t_1 = v/g \approx 0.55$ s
Read literally, time in glycerine $= 1.5 - 0.55 = 0.95$ s: $s = 5.42 \times 0.95 - \frac{1}{2} \times 2.66 \times 0.95^2 \approx 3.9$ m
The intended reading takes 1.5 s as the time inside the glycerine: $s = 5.42 \times 1.5 - \frac{1}{2} \times 2.66 \times 1.5^2 \approx 8.13 - 2.99 \approx 5.1$ m
This is nearest to 5.5 m. So the answer is (c) 5.5 m.

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