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1

AIEEE 2002

MCQ (Single Correct Answer)
The minimum velocity (in $$m{s^{ - 1}}$$) with which a car driver must traverse a flat curve of radius 150 m and coefficient of friction $$0.6$$ to avoid skidding is
A
$$60$$
B
$$30$$
C
$$15$$
D
$$25$$

Explanation

For no skidding along curved track,

The maximum velocity possible $${v_{\max }} = \sqrt {\mu rg} $$

Here $$\mu = 0.6,\,r = 150m,\,g = 9.8$$

$$\therefore$$ $${v_{\max }} = \sqrt {0.6 \times 150 \times 9.8} \simeq 30m/s$$
2

AIEEE 2002

MCQ (Single Correct Answer)
Three identical blocks of masses $$m=2$$ $$kg$$ are drawn by a force $$F=10.2$$ $$N$$ with an acceleration of $$0.6$$ $$m{s^{ - 2}}$$ on a frictionless surface, then what is the tension (in $$N$$) in the string between the blocks $$B$$ and $$C$$?
A
$$9.2$$
B
$$3.4$$
C
$$4$$
D
$$7.8$$

Explanation

Assume tension between block A and B is T1 and block B and C is T2. Acceleration a = 0.6 m/s2.

For block A :

F - T1 = ma

For block B :

T1 - T2 = ma

By adding both those equation we get,

F - T2 = 2ma

T2 = F - 2ma

T2 = 10.2 - 2$$ \times $$2$$ \times $$0.6 = 10.2 - 2.4 = 7.8 N
3

AIEEE 2002

MCQ (Single Correct Answer)
A light string passing over a smooth light pulley connects two blocks of masses $${m_1}$$ and $${m_2}$$ (vertically). If the acceleration of the system is $$g/8$$, then the ratio of the masses is
A
$$8:1$$
B
$$9:7$$
C
$$4:3$$
D
$$5:3$$

Explanation


Assume that, mass m1 is greater than mass m2, so the heavier mass m1 is accelerating downward and the lighter mass m2 is accelerating upwards.

For mass $${m_1}$$ the equation will be

$${m_1}$$$$g-T=$$$${m_1}$$$$a$$

For mass $${m_2}$$ the equation will be

$$T-$$$${m_2}$$$$g=$$$${m_2}$$$$a$$

Adding those equations we get

$$a = {{\left( {{m_1} - {m_2}} \right)g} \over {{m_1} + {m_2}}}$$

$$\therefore$$ $${g \over 8} = {{\left( {{m_1} - {m_2}} \right)g} \over {{m_1} + {m_2}}}$$

$$ \Rightarrow {1 \over 8} = {{{{{m_1}} \over {{m_2}}} - 1} \over {{{{m_1}} \over {{m_2}}} + 1}}$$

$$ \Rightarrow$$ $${{{m_1}} \over {{m_2}}} + 1 =$$ $${8\left( {{{{m_1}} \over {{m_2}}} - 1} \right)}$$

$$ \Rightarrow$$ $${{{m_1}} \over {{m_2}}} = {9 \over 7}$$

4

AIEEE 2002

MCQ (Single Correct Answer)
Two forces are such that the sum of their magnitudes is $$18$$ $$N$$ and their resultant is $$12$$ $$N$$ which is perpendicular to the smaller force. Then the magnitudes of the forces are
A
$$12N,$$ $$6N$$
B
$$13N,$$ $$5N$$
C
$$10N,$$ $$8N$$
D
$$16N$$, $$2N.$$

Explanation

Let the two forces be $${F_1}$$ and $${F_2}$$ and let $${F_2}$$ is smaller than $$ {F_1} $$ and assume $$R$$ is the resultant force.

Given $${F_1} + {F_2} = 18$$ $$\,\,\,\,\,\,$$ ....$$(i)$$

From the right angle triangle, $$F_2^2 + {R^2} = F_1^2$$

or $$F_1^2 - F_2^2 = {R^2}$$

or $$\left( {{F_1} + {F_2}} \right)$$$$\left( {{F_1} - {F_2}} \right)$$ = $${R^2}$$

or $$\left( {18} \right)\left( {{F_1} - {F_2}} \right)$$ = $${\left( {12} \right)^2}$$ = 144

or $$\left( {{F_1} - {F_2}} \right) = 8$$ $$\,\,\,\,\,\,$$ ....$$(ii)$$

By solving equation $$(i)$$ and $$(ii)$$ we get,

$${{F_1} = 13\,N}$$ and $${{F_2} = 5\,N}$$

Questions Asked from Laws of Motion

On those following papers in MCQ (Single Correct Answer)
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