Question

A magnetic field exists in a smoke chamber shown perpendicular to the plane of paper (either upwards or downwards). A beam coming out of a radioactive material [consisting of $\alpha ,{\beta }^{-},\gamma$ particles, proton $\left(H^{+}\right)$, Neutron] enters into the chamber through a fine hole, the four rays are named as $A,B,C$ and $D$.
[Assume $\alpha ,{\beta }^{-},$ Neutron and proton have nearly the same velocity.]
$\begin{array}{cccc}& \text{Column-I}& & \text{Column-II}\\ \text{(A)}& A& \text{(P)}& \alpha \\ \text{(B)}& B& \text{(Q)}& {\beta }^{-}\\ \text{(C)}& C& \text{(R)}& \gamma \\ \text{(D)}& D& \text{(S)}& H^{+}\\ & & \text{(T)}& \text{Neutron}\end{array}$
Which of the following option has the correct combination considering column-I and column-II.

• A. $A\to S;B\to P;C\to R,T;D\to Q$
• B. $A\to T;B\to Q;C\to R,T;D\to Q$
• C. $A\to S;B\to P;C\to P,Q,R,T;D\to Q$
• D. $A\to P,S;B\to P,R;C\to R,T;D\to Q$

Answer 1

The correct option is A $A\to S;B\to P;C\to R,T;D\to Q$

As magnetic field is perpendicular to the plane of paper(inwards) so when we consider right hand thumb rule, we can conclude that
$1.$ Positively charged particle will go in upward direction,
$2.$ Negatively charged particle will go in downward direction, hence (C) $\to$ R,T.
$3.$ Uncharged particle will move undeflected, hence (D) $\to$ Q

Now between $H^{+}$ and $\alpha$ particles
Radius of curvature of the curved path
$R=\frac{mv}{qB}$
Radius of $\alpha$ particle is
$R_{\alpha }=\frac{4mv}{2qB}$
Radius of $H^{+}$ is
$R_P=\frac{mv}{qB}=\frac{R_{\alpha }}{2}$
Radius of $H^{+}<$ Radius of $\alpha$ particle.

Hence (A) $\to$ S; (B) $\to$ P

Hence from all of the given options, option (A) is suitable as the correct answer.