MHT CET · Physics · Motion In One Dimension
A small steel ball is dropped from a height of \(1.5 \mathrm{~m}\) into a glycerine jar. The ball reaches the bottom of the jar 1.5 second after it was dropped. If the retardation is \(2.66 \mathrm{~m} / \mathrm{s}^2\), the height of the glycerine in the jar is about (acceleration due to gravity \(g=9.8 \mathrm{~m} / \mathrm{s}^2\) )
- A \(7.0 \mathrm{~m}\)
- B \(7.5 \mathrm{~m}\)
- C \(5.5 \mathrm{~m}\)
- D \(3.2 \mathrm{~m}\)
Answer & Solution
Correct Answer
(C) \(5.5 \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
The velocity of the ball after it has been dropped from height till it reaches the glycerine surface is
\(\mathrm{v}_{\mathrm{i}}^2=0+2 \mathrm{gh} \quad \quad \ldots .\left(\because \mathrm{v}^2-\mathrm{u}^2=2 \mathrm{gh}\right)\)
\(v_i^2=2 \times 9.8 \times 1.5\)
\(v_i^2=29.4\)
The velocity of the ball after it enters glycerine is \(v_f^2=v_i^2-2 g h\)
\(0=29.4-(2 \times 2.66 \times h)\)
\(\therefore \quad \mathrm{h}=\frac{29.4}{2 \times 2.66}=5.5 \mathrm{~m}\)
\(\mathrm{v}_{\mathrm{i}}^2=0+2 \mathrm{gh} \quad \quad \ldots .\left(\because \mathrm{v}^2-\mathrm{u}^2=2 \mathrm{gh}\right)\)
\(v_i^2=2 \times 9.8 \times 1.5\)
\(v_i^2=29.4\)
The velocity of the ball after it enters glycerine is \(v_f^2=v_i^2-2 g h\)
\(0=29.4-(2 \times 2.66 \times h)\)
\(\therefore \quad \mathrm{h}=\frac{29.4}{2 \times 2.66}=5.5 \mathrm{~m}\)
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