Category Archives: IIT JEE

A horizontal force F is applied at the center of mass …

A horizontal force F is applied at the center of mass of a cylindrical object of mass m and radius R, perpendicular to its axis as shown in the figure. The coefficient of friction between the object and the ground is μ. The center of mass of the object has an acceleration a. The acceleration due to gravity is g. Given that the object rolls without slipping, which of the following statement(s) is(are) correct?

(A) For the same F, the value of a does not depend on whether the cylinder is solid or hollow
(B) For a solid cylinder, the maximum possible value of a is 2μg
(C) The magnitude of the frictional force on the object due to the ground is always μmg
(D) For a thin-walled hollow cylinder, 𝑎=𝐹2𝑚 Continue reading A horizontal force F is applied at the center of mass …

f(x)=x23x6x2+2x+4

Domain/Range/Onto/Into

Let f:RR be defined by f(x)=x23x6x2+2x+4. Then which of the following statements is(are) correct?

(A) f is onto
(B) Range of f is [32,2]
(C) f is into
(D) Domain of f is [-4, 0] Continue reading f(x)=x23x6x2+2x+4

Domain/Range/Onto/Into

I=ππsin2x1+exdx=?

Let x=t, so dx=dt

I=ππsin2t1+etdt=ππetsin2t1+etdt=ππexsin2x1+exdx

I+I=ππsin2x1+exdx+ππexsin2x1+exdx

2I=ππ(1+ex)sin2x1+exdx=ππsin2xdx

2I=π02sin2xdx=π0(1cos2x)dx

2I=(xsin2x2)|π0=π

I=π2

I=10ln(1+x)1+x2dx=?

Let, x=tanθ

dx=sec2θdθ

I=π/40ln(1+tanθ)1+tan2θsec2θdθ

=π/40ln(sinθ+cosθcosθ)dθ

=π/40ln(sinθ+cosθ)dθπ/40lncosθdθ

=π/40ln{2(cosθ.12+sinθ.12)}dθπ/40lncosθdθ

=π/40ln{2(cosθ.cosπ4+sinθ.sinπ4)}dθπ/40lncosθdθ

=π/40ln{2cos(θπ4)}dθπ/40lncosθdθ

=π/40ln2dθ+π/40lncos(θπ4)dθπ/40lncosθdθ

Let, θπ4=ϕ for the second integral.

I=π4ln20π/4lncos(ϕ)dϕπ/40lncosθdθ

=π4ln2+π/40lncosϕdϕπ/40lncosθdθ

=π4ln2=π8ln2

A projectile is thrown from a point O ….

A projectile is thrown from a point O on the ground at an angle 45° from the vertical and with a speed 5√2 m/s. The projectile at the highest point of its trajectory splits into two equal parts. One part falls vertically down to the ground, 0.5 s after the splitting. The other part, t seconds after the splitting, falls to the ground at a distance x meters from the point O. The acceleration due to gravity g=10m/s2.

Q.1 The value of t is ___ .
Q.2 The value of x is ___ .

Solution

The part that falls vertically down to the ground has initial speed = 0. So, it falls freely taking 0.5 s to reach the ground. Continue reading A projectile is thrown from a point O ….

In the circuit shown, the switch S is connected to …

In the circuit shown, the switch S is connected to position P for a long time so that the charge on the capacitor becomes q1 μC. Then S is switched to position Q. After a long time, the charge on the capacitor is q2 μC.

Q.1 The magnitude of q1 is ___ .
Q.2 The magnitude of q2 is ___ .

Solution

The figure below shows the situation when switch S is connected to position P.

No current flows through the capacitor in steady state. Continue reading In the circuit shown, the switch S is connected to …