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Standard XII

Physics

Rotational Work and Energy

A circular ri...

Question

A circular ring of radius $3 c m$ is suspended horizontally from a point $4 c m$ vertically above the center by $4$ strings attached at equal intervals to its circumference.If the angle between two consecutive strings be $θ$ then

\frac{4}{5}

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\frac{4}{25}

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\frac{16}{25}

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Solution

The correct option is C $\frac{16}{25}$
Clearly, $A C = \sqrt{(3^{2} + 4^{2})} = 5 = A D$ and $∠ C B D = 90$
So, $C D = \sqrt{(3^{2} + 3^{2})} = 3 \sqrt{2}$
$∴ cos θ = \frac{5^{2} + 5^{2} - {(3 \sqrt{2})}^{2}}{2 \times 5 \times 5} = \frac{16}{25}$

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A circular ring of radius 3 cm is suspended horizontally from a point 4 cm vertically above the center by 4 strings attached at equal intervals to its circumference.If the angle between two consecutive strings be θ then

The correct option is C 1625Clearly, AC=√(32+42)=5=AD and ∠CBD=90So, CD=√(32+32)=3√2∴cosθ=52+52−(3√2)22×5×5=1625

A circular ring of radius $3 c m$ is suspended horizontally from a point $4 c m$ vertically above the center by $4$ strings attached at equal intervals to its circumference.If the angle between two consecutive strings be $θ$ then

The correct option is C $\frac{16}{25}$
Clearly, $A C = \sqrt{(3^{2} + 4^{2})} = 5 = A D$ and $∠ C B D = 90$
So, $C D = \sqrt{(3^{2} + 3^{2})} = 3 \sqrt{2}$
$∴ cos θ = \frac{5^{2} + 5^{2} - {(3 \sqrt{2})}^{2}}{2 \times 5 \times 5} = \frac{16}{25}$