my and Tadiany28.A block of of mass m, =4=Ikg another mass m, =2 kg, are placed together (see figure) on an inclined planewith angle of inclination 9. Various values of A are given in List-I. The coefficient of friction between theblock m, and the plane is always zero The.coefficient of static and dynamic friction between the blocķ m,and the plane are equal to u =0.3: 1Tn List-T expressiöns for the frictioni on block m, are given. Matchthe coirect expression of the friction in Lişt-II with the angles given in List-I, and the choose the correctoption. The acceleration due to gravity is denoted by g.[Useful inforınation : tantanList-IList-I(Р) 0 — 5°(1) m2g sine(Q) 0 = 10°(2) (m +m2)gsin 0(R) 0 = 15°(3) um,g cos 0(S) 0 = 20°:?(4) µ(m, + m2)g cose(а) Р-1, Q-1, R-1, S-3(с) Р-2, Q-2, R-2, S-4(b) Р-2, Q-2, R-2, S-3(d) Р-2, Q-2, R-3, S-3

The correct option from the options below, given,

A block of of mass m, = 1 kg another mass m, = 2 kg, are placed together (see figure) on an inclined plane with angle of inclination g. Various vahies of e are given in List-I. The coefficient of friction between the block .... and the plane is always zero The coefficiënt of static and dynamric friction between the block m, • and the plane are equal to u =0.3. In List-U expressións for the frictiori on block m, are given. Match the coirect expression of the friction in Lişt-II with the angles given in List-I, and the choose the correct option. The acceleration due to gravity is denoted by g. (Useful information : 1an(s ºal; tan(11.5Y=02; tan(16.5°}=0.3] List-I List-I (P) 0 = 5° (1) m,g sin e m. (Q) 0 =10° (2) (m, +m2)g sin 0 (R) 0 = 15° (3) um,g cos 0 (S) 0 = 20°: (4) м (т + т, m2)g cose

A block of of mass m, = 1 kg another mass m, = 2 kg, are placed together (see figure) on an inclined plane with angle of inclination g. Various vahies of e are given in List-I. The coefficient of friction between the block .... and the plane is always zero The coefficiënt of static and dynamric friction between the block m, • and the plane are equal to u =0.3. In List-U expressións for the frictiori on block m, are given. Match the coirect expression of the friction in Lişt-II with the angles given in List-I, and the choose the correct option. The acceleration due to gravity is denoted by g. (Useful information : 1an(s ºal; tan(11.5Y=02; tan(16.5°}=0.3] List-I List-I (P) 0 = 5° (1) m,g sin e m. (Q) 0 =10° (2) (m, +m2)g sin 0 (R) 0 = 15° (3) um,g cos 0 (S) 0 = 20°: (4) м (т + т, m2)g cose

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Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Concept explainers

Rotational Equilibrium And Rotational Dynamics

In physics, the state of balance between the forces and the dynamics of motion is called the equilibrium state. The balance between various forces acting on a system in a rotational motion is called rotational equilibrium or rotational dynamics.

Equilibrium of Forces

The tension created on one body during push or pull is known as force.

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my and Tadiany
28.
A block of of mass m, =
4=Ikg another mass m, =2 kg, are placed together (see figure) on an inclined plane
with angle of inclination 9. Various values of A are given in List-I. The coefficient of friction between the
block m, and the plane is always zero The.coefficient of static and dynamic friction between the blocķ m,
and the plane are equal to u =0.3: 1Tn List-T expressiöns for the frictioni on block m, are given. Match
the coirect expression of the friction in Lişt-II with the angles given in List-I, and the choose the correct
option. The acceleration due to gravity is denoted by g.
[Useful inforınation : tan
tan
List-I
List-I
(Р) 0 — 5°
(1) m2g sine
(Q) 0 = 10°
(2) (m +m2)gsin 0
(R) 0 = 15°
(3) um,g cos 0
(S) 0 = 20°:?
(4) µ(m, + m2)g cose
(а) Р-1, Q-1, R-1, S-3
(с) Р-2, Q-2, R-2, S-4
(b) Р-2, Q-2, R-2, S-3
(d) Р-2, Q-2, R-3, S-3

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