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JEE Mains Previous Years Questions with Solutions

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1

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.
2

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 $$(2)$$
$${{{F_e}} \over {mg}} = {{2{F_e}'} \over {mg}} \Rightarrow {F_e} = 2{F_e}' \Rightarrow {F_e}' = {{{F_e}} \over 2}$$

3

AIEEE 2009

MCQ (Single Correct Answer)
Two wires are made of the same material and have the same volume. However wire $$1$$ has cross-sectional area $$A$$ and wire $$2$$ has cross-sectional area $$3A.$$ If the length of wire $$1$$ increases by $$\Delta x$$ on applying force $$F,$$ how much force is needed to stretch wire $$2$$ by the same amount?
A
$$4F$$
B
$$6F$$
C
$$9F$$
D
$$F$$

Explanation


As shown in the figure, the wires will have the same Young's modulus (same material) and the length of the wire of area of cross-section $$3A$$ will be $$\ell /3$$ (same volume as wire $$1$$).
For wire $$1,$$
$$y = {{F/A} \over {\Delta x/\ell }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$$
For wire $$2.$$
$$Y = {{F'/3A} \over {\Delta x/\left( {\ell /3} \right)}}........(ii)$$
From $$(i)$$ and $$(ii),$$ $${F \over A} \times {\ell \over {\Delta x}} = {{F'} \over {3A}} \times {\ell \over {3\Delta x}} \Rightarrow F' = 9F$$
4

AIEEE 2008

MCQ (Single Correct Answer)
A jar is filled with two non-mixing liquids $$1$$ and $$2$$ having densities $${\rho _1}$$ and $${\rho _2}$$ respectively. A solid ball, made of a material of density $${\rho _3}$$, is dropped in the jar. It comes to equilibrium in the position shown in the figure. Which of the following is true for $${\rho _1}$$ , $${\rho _1}$$ and $${\rho _3}$$ ?
A
$${\rho _3} < {\rho _1} < \rho {}_2$$
B
$${\rho _1} > {\rho _3} > \rho {}_2$$
C
$${\rho _1} < {\rho _2} < \rho {}_3$$
D
$${\rho _1} < {\rho _3} < \rho {}_2$$

Explanation

From the figure it is clear that liquid $$1$$ floats on liquid $$2.$$ The lighter liquid floats over heavier liquid. Therefore we can conclude that $${\rho _1} < {\rho _2}$$

Also $${\rho _3} < {\rho _2}$$ otherwise the ball would have sink to the bottom of the jar.

Also $${\rho _3} > {\rho _1}$$ otherwise the ball would have floated in liquid $$1.$$ From the above discussion we conclude that $${\rho _1} < {\rho _2} < {\rho _3}.$$

Questions Asked from Properties of Matter

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