JEE Advanced 2026 · Paper 2, Physics question 6

Here $\frac qm=\frac{10^{-6}}{10^{-6}}=1$ C/kg.

$0<t<0.2$ s. $a_x=\frac qmE=1$, $a_y=-10$:

• $y=2t-5t^2$, so $y(0.2)=0.2$ m and $v_y(0.2)=0$;
• $v_x(0.2)=1.2$ m/s.

$t>0.2$ s. $\vec B=6\hat\jmath$ is perpendicular to $\vec v=1.2\hat\imath$, so the particle circles in a horizontal plane with radius
$$r=\frac{mv}{qB}=\frac{1.2}{6}=0.2\text{ m}=20\text{ cm}.$$
(C) true.

Gravity is parallel to $\vec B$, so the vertical motion is free fall from rest: $y=0.2-5(t-0.2)^2$.

• $t=0.3$: $y=0.2-0.05=0.15$ m $=15$ cm. (A) true.
• $t=0.4$: $y=0$, not 10 cm. (B) false.
• $y=0$ first at $t=0.4$ s, not $0.35$ s. (D) false.