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JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2011

MCQ (Single Correct Answer)
Water is flowing continuously from a tap having an internal diameter $$8 \times {10^{ - 3}}\,\,m.$$ The water velocity as it leaves the tap is $$0.4\,\,m{s^{ - 1}}$$ . The diameter of the water stream at a distance $$2 \times {10^{ - 1}}\,\,m$$ below the tap is close to :
A
$$7.5 \times {10^{ - 3}}m$$
B
$$9.6 \times {10^{ - 3}}m$$
C
$$3.6 \times {10^{ - 3}}m$$
D
$$5.0 \times {10^{ - 3}}m$$

Explanation

From Bernoulli's theorem,
$${P_0} + {1 \over 2}\rho v_1^2\rho gh = {P_0} + {1 \over 2}\rho v_2^2 + 0$$
$${v_2} = \sqrt {v_1^2 + 2gh} $$
$$ = \sqrt {0.16 + 2 \times 10 \times 0.2} $$
$$ = 2.03\,m/s$$
From equation of continuity
$${A_2}{v_2} = {A_1}{v_1}$$
$$\pi {{D_2^2} \over 4} \times {v_2} = \pi {{D_1^2} \over 4}{v_1}$$
$$ \Rightarrow \,\,{D_1} = {D_2}\sqrt {{{{v_1}} \over {{v_2}}}} = 3.55 \times {10^{ - 3}}m$$
2

AIEEE 2011

MCQ (Single Correct Answer)
Work done in increasing the size of a soap bubble from a radius of $$3$$ $$cm$$ to $$5$$ $$cm$$ is nearly (Surface tension of soap solution $$ = 0.03N{m^{ - 1}},$$
A
$$0.2\pi mJ$$
B
$$2\pi mJ$$
C
$$0.4\pi mJ$$
D
$$4\pi mJ$$

Explanation

$$W = T \times \,\,$$ change in surface area
$$W = 2T4\pi \left[ {{{\left( 5 \right)}^2} - {{\left( 3 \right)}^2}} \right] \times {10^{ - 4}}$$
$$ = 2 \times 0.03 \times 4\pi \left[ {25 - 9} \right] \times {10^{ - 4}}\,J$$
$$ = 0.4\pi \times {10^{ - 3}}\,J$$
$$ = 0.4\pi mJ$$
3

AIEEE 2010

MCQ (Single Correct Answer)
A ball is made of a material of density $$\rho $$ where $${\rho _{oil}}\, < \rho < {\rho _{water}}$$ with $${\rho _{oil}}$$ and $${\rho _{water}}$$ representing the densities of oil and water, respectively. The oil and water are immiscible. If the above ball is in equilibrium in a mixture of this oil and water, which of the following pictures represents its equilibrium position?
A
B
C
D

Explanation

Oil will float on water so, $$(2)$$ or $$(4)$$ is the correct option, But density of ball is more than that of oil, hence it will sinkin oil.
4

AIEEE 2010

MCQ (Single Correct Answer)
Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle of $${30^ \circ }$$ with each other. When suspended in a liquid of density $$0.8g$$ $$c{m^{ - 3}},$$ the angle remains the same. If density of the material of the sphere is $$1.6$$ $$g$$ $$c{m^{ - 3}},$$ the dielectric constant of the liquid is
A
$$4$$
B
$$3$$
C
$$2$$
D
$$1$$

Explanation


$${F_e} = T\sin {15^ \circ }\,\,;$$
$$mg = T\cos {15^ \circ }$$
$$ \Rightarrow \tan {15^ \circ } = {{{F_e}} \over {mg}}$$ $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$$
In liquid, $${F_e}' = T'\sin {15^ \circ }$$ $$\,\,\,\,\,\,...(ii)$$


$$mg = {F_B} + T'\cos {15^ \circ }$$
$${F_B}' = V\left( {d - \rho } \right)g = V\left( {1.6 - 0.8} \right)g = 0.8\,Vg$$
$$ = 0.8{m \over d}g = {{0.8mg} \over {1.6}} = {{mg} \over 2}$$
$$\therefore$$ $$mg = {{mg} \over 2} + T'\cos {15^ \circ }$$
$$ \Rightarrow {{mg} \over 2} = T'\cos {15^ \circ }$$ $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( B \right)$$
From $$(A)$$ and $$(B),$$ $$\tan \,{15^ \circ } = {{2{F_e}'} \over {mg}}\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$
From $$(1)$$ and $$(2)$$
$${{{F_e}} \over {mg}} = {{2{F_e}'} \over {mg}} \Rightarrow {F_e} = 2{F_e}' \Rightarrow {F_e}' = {{{F_e}} \over 2}$$

Questions Asked from Properties of Matter

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