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Physics · LAWS OF MOTION
Englishएक हल्की डोरी द्वारा छत से लटकाए गए \( 0.1\text{ kg} \) द्रव्यमान के एक छोटे गोले को क्षैतिज बल \( F \) द्वारा तब तक खींचा जाता है जब तक कि डोरी साम्यावस्था में ऊर्ध्वाधर से \( 45^\circ \) का कोण न बना ले। \( g = 10\text{ m/s}^2 \) लेते हुए, क्षैतिज बल \( F \) और डोरी के तनाव \( T \) का परिमाण क्रमशः होगा:
सही उत्तर · विकल्प 3
\( F = 1 N, T = \sqrt{2} N \)
व्याख्या
In equilibrium, the forces acting on the sphere are the gravitational force \( mg \) acting downwards, the horizontal force \( F \) acting horizontally, and the tension \( T \) in the string acting at an angle of \( 45^\circ \) to the vertical.
Resolving forces vertically: \( T \cos 45^\circ = mg \)
Given \( m = 0.1\text{ kg} \) and \( g = 10\text{ m/s}^2 \), we have:
\( T \left(\frac{1}{\sqrt{2}}\right) = (0.1)(10) = 1\text{ N} \implies T = \sqrt{2}\text{ N} \)
Resolving forces horizontally: \( F = T \sin 45^\circ \)
\( F = (\sqrt{2}) \left(\frac{1}{\sqrt{2}}\right) = 1\text{ N} \)
Therefore, \( F = 1\text{ N} \) and \( T = \sqrt{2}\text{ N} \).
Resolving forces vertically: \( T \cos 45^\circ = mg \)
Given \( m = 0.1\text{ kg} \) and \( g = 10\text{ m/s}^2 \), we have:
\( T \left(\frac{1}{\sqrt{2}}\right) = (0.1)(10) = 1\text{ N} \implies T = \sqrt{2}\text{ N} \)
Resolving forces horizontally: \( F = T \sin 45^\circ \)
\( F = (\sqrt{2}) \left(\frac{1}{\sqrt{2}}\right) = 1\text{ N} \)
Therefore, \( F = 1\text{ N} \) and \( T = \sqrt{2}\text{ N} \).
