A body is sliding down an inclined plane having coefficient of friction $0.5$. If the normal reaction is twice that of the resultant downward force along the incline, the angle between the inclined plane and the horizontal is ....... $^o$
A$15$
B$30$
C$45$
D$60$
Medium
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C$45$
c (c) Resultant downward force along the incline $ = mg(\sin \theta - \mu \cos \theta )$
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A uniform wooden stick of mass $1.6 \mathrm{~kg}$ and length $l$ rests in an inclined manner on a smooth, vertical wall of height $h( < l)$ such that a small portion of the stick extends beyond the wall. The reaction force of the wall on the stick is perpendicular to the stick. The stick makes an angle of $30^{\circ}$ with the wall and the bottom of the stick is on a rough focr. The reaction of the wall on the stick is equal in magnitude to the reaction of the floor on the st $ck$. The ratio $h / l$ and the frictional force $f$ at the bottom of the stick are $\left(g=10 \mathrm{~m} \mathrm{~s}^{-2}\right)$
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$Assertion$ : There is a stage when frictional force is not needed at all to provide the necessary centripetal force on a banked road.
$Reason$ : On a banked road, due to its inclination the vehicle tends to remain inwards without any chances of skidding.
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As shown in the figure, block m and wedge $M$ move together with a horizontal acceleration of $20\, m/s^2$. Given $m = 1\, kg$, $\mu = 0.6$ (between $m$ and $M$) and $g = 10\, m/s^2$. Choose the $CORRECT$ alternative :-
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