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Grade: 12
`        A particle subjected to two equal forces along two different directions. If one of the forces is halved, the angle which the resultant makes with th other is also halved. The angle between the forces is-(a) 45    (b) 60      (c)90     (d) 120`
4 years ago

Answers : (2)

Vasantha Kumari
askIITians Faculty
38 Points
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4 years ago
Arun Kumar
IIT Delhi
askIITians Faculty
256 Points
```							Hi$\\\vec f_1+\vec f_2=\vec f_{r1} \\\vec f_1+{\vec f_2 \over 2}=\vec f_{r2} \\\vec f_{r1}.\vec f_1=|\vec f_{r1}||\vec f_1|cos\theta \\\vec f_{r2}.\vec f_1=|\vec f_{r2}||\vec f_1 |cos{\theta \over 2} \\\vec f_{r1}-\vec f_{r2}={\vec f_2 \over 2} \\\vec f_{r2}.\vec f_1+{\vec f_2 \over 2}.\vec f_1=|\vec f_{r1}||\vec f_1|cos\theta \\|\vec f_{r2}||\vec f_1 |cos{\theta \over 2}+{\vec f_2 \over 2}.\vec f_1=|\vec f_{r1}||\vec f_1|cos\theta$$\\|\vec f_{r2}||\vec f_1 |cos{\theta \over 2}+{|\vec f_2| \over 2}.|\vec f_1|cos\theta_{req}=|\vec f_{r1}||\vec f_1|cos\theta \\since |\vec f_1=|\vec f_2| \\=>|\vec f_{r2}|cos{\theta \over 2}+{|\vec f_2| \over 2}cos\theta_{req}=|\vec f_{r1}|cos\theta \\=>also since |\vec f_1=|\vec f_2|=|f|  so \theta_{req}=2\theta \\|\vec f_{r1}|^2=2|f|^2+2|f|^2cos(2\theta) \\|\vec f_{r2}|^2=5/4|f|^2+|f|^2cos(2\theta) \\=>\sqrt{5/4+cos(2\theta)}cos{\theta \over 2}+{cos\theta_{req} \over 2}=\sqrt{2+2cos(2\theta)}cos\theta \\=>\sqrt{5/4+cos(2\theta)}cos{\theta \over 2}+{cos2\theta \over 2}=\sqrt{2+2cos(2\theta)}cos\theta \\=>\sqrt{5/4+cos(2\theta)}cos{\theta \over 2}+{cos2\theta \over 2}=\sqrt{2+2cos(2\theta)}cos\theta$See for a equation like this you gotta use the options given.start putting the values in the equation.we find$\theta_{req}=120$=>angle between the forces =120Thanks & Regards, Arun Kumar, Btech,IIT Delhi, Askiitians Faculty
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4 years ago
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