IIT JEE Main 2016 Physics Question 14

IIT JEE Main 2016 Physics Question No 14

A point particle of mass m, moves along the uniformly rough track PQR as shown in the figure. The coefficient of friction, between the particle and the rough track equals µ . The particle is released, from rest, from the point P and it comes to rest at a point R. The energies, lost by the ball, over the parts, PQ and QR, of the track, are equal to each other, and no energy is lost when particle changes direction from PQ to QR. The values of the coefficient of friction µ and the distance x(= QR), are respectively close to

IIT JEE Main 2016 Physics Question No 14

(1) 0.29 and 3.5 m
(2) 0.29 and 6.5 m
(3) 0.2 and 6.5 m
(4) 0.2 and 3.5 m

Solution

PQ = (h/sin 30º) = 2h = 4m
Given that work done against friction along track PQ and QR are equal.
therefore, µ mg cos 30º  x 4 =  µ mg x
(4√3)/2=x
=> x = 2√3 = 2 x 1.732 = 3.46 ≅ 3.5 m
Loss in PE = work done against friction along PQ and QR

hence, mg 2 = µ mg( (√3)/2)x4 + µ mg (3.5)

So we get µ = 0.29

Answer

Correct answer to given problem No 14 of IIT JEE Main Physics paper of 2016 is µ = 0.29 and QR = 3.5 i.e. option number 1.

IIT JEE Main 2016 Physics Question No 10

IIT JEE Main 2016 Physics Question No 10

The region between two concentric spheres of radii ‘a’ and ‘b’, respectively (see figure ), have volume charge density ρ = A/r where A is a constant and r is the distance from the centre. At the centre of the spheres is a point charge Q. The value of A such that the electric field in the region between the spheres will be constant is:

IIT JEE Main 2016 Physics Question No 10
(1) 2Q/(Π(a2-b2))
(2) 2Q/(Π(α2)
(3) Q/(2Π(α2)
(4) Q/(2Π(b2-a2))

Solution

Using Gauss Theorem for a Gaussian surface for radius r, we have

E 4Πr2 = (1/ε0)[Q + ∫ar(A/r)4Πr2dr]
E = (Q/(4Πε0r2)) + [(1/(4Πε0r2))2ΠA{r2 – ar2}]
E = (Q/(4Πε0r2)) + (A/2ε0) – (Aa2/2r2ε0)

for E to be independent of r

(Q/(4Πε0r2)) – (Aa2/2r2ε0) = 0

=> A = Q/2Πa2

Answer

Correct answer to given problem No 10 of IIT JEE Main Physics paper of 2016 is A = Q/2Πa2 i.e. option number 3.

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