11) work done in the whole process is equal to Wa,0 + Wa,a
now dWa,0 = -K(a\hat{i}+0\hat{j}).dr \Rightarrow W = -K\int(a\hat{i}+0\hat{j}){dr}\hat{i}=-Ka^{2}/2
similarly for 2nd part
hence net work done is -Ka2/2 + ( - Ka2/2 ) = - Ka2
1) A force F=-K((x i + y j) acts on a particle moving in the x-y plane.(K is a positive constant).Starting from origin if the particle is taken to (a,0) along x-axis and from there to (a,a) parallel to y-axis,the work done y the force is...
a)-2Ka2 b)-Ka2 c)2Ka2 d)N.O.T
the ans i was getting is a but the ans given is b
2) A chain of mass m is held so that its bottom end just touches the table.Then it is allowed fall freely .What i the net force on the table?
a)mg b)2mg c)3mg d)N.O.T
ans is c
3)A current carrying conductor which is of the form of a sine wave given as y = a sinx lies on a magnetic field. What is the force exerted by the magnetic field??
a)iB b)iB(pi) c)2iB(pi) d) data insufficient
the ans is c
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4 Answers
112) let length of the chain is L
since chain is allowed to free fall velocity acquired v = √2gL
weight = mg ( downward )
thrust force is given by Ft = vr dmdt where vr = relative velocity = v,
and dmdt = λv ( λ = mass per unit length = mL )
hence Ft = λv2 = mL x 2gL = 2mg
hence net force = Ft + W = 2mg + mg = 3mg
hence required answer is 3mg . So option (c) is correct [1]
11@manmay thats what i had also done...but the solution says..
"In the first part F=-K(xi+0j) = -Kxi dr = dx i ∫F.dr = -Ka2/2 . Similarly for the 2nd part......"
i couldnt get what the solution is trying to tell....
13) since sine wave has a peroid of 2Ï€ so force will be repeated after every 2Ï€ cycle
hence we will take force for one cycle
force per unit length ( F/l ) hence = B i ( 2Ï€) ,,,,,,,,,,,,, which is (c)