Question

In below figure two blocks are placed on fixed inclined plane and friction coefficient between the blocks and inclined plane are ${\mu }_1$ and ${\mu }_2$ respectively as shown in figure. Value of ${\mu }_1$ and ${\mu }_2$ are given in column 1 and value of $a_1$ and $a_2$ are given in column 2 and column 3 respectively.

$\begin{array}{cccccc}& \text{Column 1}& & \text{Column 2}& & \text{Column 3}\\ \left(i\right)& {\mu }_1=0.8& \left(i\right)& a_1=2m/s^2& \left(P\right)& a_2=2m/s^2\\ & {\mu }_2=0.8& & & & \\ \left(ii\right)& {\mu }_1=0.5& \left(ii\right)& a_1=0.8m/s^2& \left(Q\right)& a_2=0.8m/s^2\\ & {\mu }_2=0.5& & & & \\ \left(iii\right)& {\mu }_1=0.5& \left(ii\right)& a_1=4.4m/s^2& \left(R\right)& a_2=4.4m/s^2\\ & {\mu }_2=0.2& & & & \\ \left(iv\right)& {\mu }_1=0.5& \left(iv\right)& a_1=0& \left(S\right)& a_2=0\\ & {\mu }_2=0.8& & & & \end{array}$

In which of the following combination both blocks will move together and normal force between them is non-zero?

• A. (II) (i) (P)
• B. (III) (ii) (Q)
• C. (I) (ii) (Q)
• D. (IV) (ii) (Q)

Answer 1

The correct option is D

(IV) (ii) (Q)

(I) As ${\mu }_1$ and ${\mu }_2>tan\theta$

$\therefore$ Both block will remain at rest

Hence, $a_1=a_2=0$ and $N=0$

(II) $a_1=a_2=g(sin{37}^{\circ }-\mu cos{37}^{\circ })$

$\Rightarrow a_1=a_2=2m/s^2$ and $N=0$

(III) $a_1=g(sin{37}^{\circ }–{\mu }_{1}cos{37}^{\circ })$

$a_1=2m/s^2$

$a_2=g(sin{37}^{\circ }-{\mu }_{2}cos{37}^{\circ })$

$a_2=4.4m/s^2$ and $N=0$

(IV) $a_1=a_2=\frac{4gsin{37}^{\circ }-{\mu }_{1}2gcos{37}^{\circ }-{\mu }_{2}2gcos{37}^{\circ }}{4}$

$\Rightarrow a_1=a_2=0.8m/s^2$ and $N\ne 0$