A solid sphere of radius 10 cm is rotating about an axis which is at a distance 15 cm from its centre. The radius of gyration about this axis is cm. The value

Question

A solid sphere of radius 10 cm is rotating about an axis which is at a distance 15 cm from its centre. The radius of gyration about this axis is $\sqrt{n}$ cm. The value of n is _____. [2026]

Smart Achievers · Accepted Answer

(265)

$Let radius of gyration is k$

$\Rightarrow m k^{2} = \frac{2}{3} m R^{2} + m d^{2}$

$k^{2} = \frac{2}{3} \times 10^{2} + 15^{2} = 265$

${(\sqrt{n})}^{2} = 265 \Rightarrow n = 265$

Solution

Q.
A solid sphere of radius 10 cm is rotating about an axis which is at a distance 15 cm from its centre. The radius of gyration about this axis is $\sqrt{n}$ cm. The value of n is _____. [2026]

Ans.
(265)

$Let radius of gyration is k$

$\Rightarrow m k^{2} = \frac{2}{3} m R^{2} + m d^{2}$

$k^{2} = \frac{2}{3} \times 10^{2} + 15^{2} = 265$

${(\sqrt{n})}^{2} = 265 \Rightarrow n = 265$

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Solution

Q. A solid sphere of radius 10 cm is rotating about an axis which is at a distance 15 cm from its centre. The radius of gyration about this axis is n cm. The value of n is _____. [2026]

Ans. (265) Let radius of gyration is k ⇒mk2=23mR2+md2 k2=23×102+152=265 (n)2=265⇒n=265

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Q.
A solid sphere of radius 10 cm is rotating about an axis which is at a distance 15 cm from its centre. The radius of gyration about this axis is $\sqrt{n}$ cm. The value of n is _____. [2026]

Ans.
(265)

$Let radius of gyration is k$

$\Rightarrow m k^{2} = \frac{2}{3} m R^{2} + m d^{2}$

$k^{2} = \frac{2}{3} \times 10^{2} + 15^{2} = 265$

${(\sqrt{n})}^{2} = 265 \Rightarrow n = 265$