JEE Main 2016 · 3 Apr · Shift 1 · Physics Q2

PhysicsSingle correct+4 / -1Question 2 of 90

A particle of mass $m$ is moving along the side of a square of side ' $a$ ', with a uniform speed $v$ in the $x$ - $y$ plane as shown in the figure :

Which of the following statements is false for the angular momentum $\vec L$ about the origin ?

\vec L=-\dfrac{mv}{\sqrt2}R\,\hat k

when the particle is moving from A to B.

\vec L=mv\left[\dfrac R{\sqrt2}-a\right]\hat k

when the particle is moving from C to D.

\vec L=mv\left[\dfrac R{\sqrt2}+a\right]\hat k

when the particle is moving from B to C.

\vec L=\dfrac{mv}{\sqrt2}R\,\hat k

when the particle is moving from D to A.

Answer and worked solution

Official answer: B

Put A at $\left(\frac R{\sqrt2},\frac R{\sqrt2}\right)$ ; the particle goes A → B → C → D, and $L_z=m(xv_y-yv_x)$ .

A → B ( $v_x=v$ , $y=\frac R{\sqrt2}$ ): $L=-\frac{mvR}{\sqrt2}\hat k$ . True.
B → C ( $v_y=v$ , $x=\frac R{\sqrt2}+a$ ): $L=mv\left(\frac R{\sqrt2}+a\right)\hat k$ . True.
C → D ( $v_x=-v$ , $y=\frac R{\sqrt2}+a$ ): $L=mv\left(\frac R{\sqrt2}+a\right)\hat k$ , not $\frac R{\sqrt2}-a$ . So (2) is false.
D → A ( $v_y=-v$ , $x=\frac R{\sqrt2}$ ): $L=-\frac{mvR}{\sqrt2}\hat k$ , so (4) as printed has the wrong sign as well.

The intended answer is (2); since (4) is also false as printed, this derived key accepts both.

JEE Main 2016 · 3 Apr · Shift 1, Physics question 2