JEE Mains · Physics · STD 11 - 4.1 newtons laws of motion
Three masses \(M =100\,kg , m _{1}=10\,kg\) and \(m_{2}=20\,kg\) are arranged in a system as shown in figure. All the surfaces are frictionless and strings are inextensible and weightless. The pulleys are also weightless and frictionless. \(A\) force \(F\) is applied on the system so that the mass \(m_{2}\) moves upward with an acceleration of \(2\,ms ^{-2}\). The value of \(F\) is \(......N\) \(\left(\right.\) Take \(\left.g =10\,ms ^{-2}\right)\)
- A \(3360\)
- B \(3380\)
- C \(3120\)
- D \(3240\)
Answer & Solution
Correct Answer
(A) \(3360\)
Step-by-step Solution
Detailed explanation
Let acceleration of \(100 kg\) block \(=a_{1}\) FBD of \(100\,kg\) block w.r.t ground \(F - T - N _{1}=100 a _{1} \ldots \ldots . .( i )\) \(FBD\) of \(20\) block wrt \(100\,kg\) \(T -20\,g =20(2)\) \(T =240\) \(N _{1}=20 a _{1}\) FBD of \(10\,kg\) block wrt \(100\,kg\)…
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