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!!Hello friends and elderbrothers!!
Class => 11
Subject => Physics
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Questions are in the attachment.
✔️All three ques. ✔️
Terms and conditions : full expalanation please it is my request.
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Jai bajarang bali
@DrS.M.
Answers
Hey there !
Solution:
42 ) Since the lower wire has uniform mass, the tension would also be constant at the middle of the wire.
Since the object is moving upwards, the mass of the string would be 0.2 * 0.5 = 0.1 kg
According to the FBD of the lower block, we would get,
Tension = ( m.g ) + ( m.a )
=> Tension = ( mg + ma ) => m ( g + a )
=> Tension = ( 1.9 + 0.1 kg ) ( 9.8 m/s² + 0.2 m/s² )
=> Tension = ( 2 ) ( 10 ) = 20 N
Hence option ( b ) is correct
43 ) Refer attachment for clear understanding.
Tension in the string would be only mg as there is zero acceleration.
Resolving Tension into components we get,
2 T cosФ = √2 mg
T = mg. So,
=> 2 mg Cos Ф = √2 mg
mg gets cancelled on both sides, we get,
=> Cos Ф = √2 / 2
=> Cos Ф = 1 / √2
=> Ф = Cos⁻¹ ( 1 / √2 )
=> Ф = 45°
Hence Option ( c ) is correct.
44 ) Resolving T into components we get,
T CosФ = W
=> T = W / CosФ => Equation 1
T Sin Ф = R => Equation 2
Substituting Equation 1 in Equation 2 we get,
=> W Sin Ф / Cos Ф = R
=> R = W Tan Ф
So this is true which implies that Statement ( d ) is true.
But only Option ( a ) has no Statement ( d ) . Hence Option ( a ) is the right answer.
Hope my answer helped !
