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JEE Mains Previous Years Questions with Solutions

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1

JEE Main 2021 (Online) 31st August Evening Shift

MCQ (Single Correct Answer)
Statement I :

Two forces $$\left( {\overrightarrow P + \overrightarrow Q } \right)$$ and $$\left( {\overrightarrow P - \overrightarrow Q } \right)$$ where $$\overrightarrow P \bot \overrightarrow Q $$, when act at an angle $$\theta$$1 to each other, the magnitude of their resultant is $$\sqrt {3({P^2} + {Q^2})} $$, when they act at an angle $$\theta$$2, the magnitude of their resultant becomes $$\sqrt {2({P^2} + {Q^2})} $$. This is possible only when $${\theta _1} < {\theta _2}$$.

Statement II :

In the situation given above.

$$\theta$$1 = 60$$^\circ$$ and $$\theta$$2 = 90$$^\circ$$

In the light of the above statements, choose the most appropriate answer from the options given below :-
A
Statement I is false but Statement II is true
B
Both Statement I and Statement II are true
C
Statement I is true but Statement II is false
D
Both Statement I and Statement II are false.

Explanation

$$\overrightarrow A = \overrightarrow P + \overrightarrow Q $$

$$\overrightarrow B = \overrightarrow P - \overrightarrow Q $$

$$\overrightarrow P \bot \overrightarrow Q $$

$$\left| {\overrightarrow A } \right| = \left| {\overrightarrow B } \right| = \sqrt {2({P^2} + {Q^2})(1 + \cos \theta )} $$

For $$\left| {\overrightarrow A + \overrightarrow B } \right| = \sqrt {3({P^2} + {Q^2})} $$

$${\theta _1} = 60^\circ $$

For $$\left| {\overrightarrow A + \overrightarrow B } \right| = \sqrt {2({P^2} + {Q^2})} $$

$${\theta _2} = 90^\circ $$
2

JEE Main 2021 (Online) 27th August Morning Shift

MCQ (Single Correct Answer)
The resultant of these forces $$\overrightarrow {OP} ,\overrightarrow {OQ} ,\overrightarrow {OR} ,\overrightarrow {OS} $$ and $$\overrightarrow {OT} $$ is approximately .......... N.

[Take $$\sqrt 3 = 1.7$$, $$\sqrt 2 = 1.4$$ Given $$\widehat i$$ and $$\widehat j$$ unit vectors along x, y axis]

A
$$9.25\widehat i + 5\widehat j$$
B
$$3\widehat i + 15\widehat j$$
C
$$2.5\widehat i - 14.5\widehat j$$
D
$$ - 1.5\widehat i - 15.5\widehat j$$

Explanation


$$\overrightarrow {{F_x}} = \left( {10 \times {{\sqrt 3 } \over 2} + 20\left( {{1 \over 2}} \right) + 20\left( {{1 \over {\sqrt 2 }}} \right) - 15\left( {{1 \over {\sqrt 2 }}} \right) - 15\left( {{{\sqrt 3 } \over 2}} \right)} \right)\widehat i$$

$$ = 9.25\widehat i$$

$$\overrightarrow {{F_y}} = \left( {15\left( {{1 \over 2}} \right) + 20\left( {{{\sqrt 3 } \over 2}} \right) + 10\left( {{1 \over 2}} \right) - 15\left( {{1 \over {\sqrt 2 }}} \right) - 20\left( {{1 \over {\sqrt 2 }}} \right)} \right)\widehat j$$

$$ = 5\widehat j$$
3

JEE Main 2021 (Online) 26th August Evening Shift

MCQ (Single Correct Answer)
The angle between vector $$\left( {\overrightarrow A } \right)$$ and $$\left( {\overrightarrow A - \overrightarrow B } \right)$$ is :

A
$${\tan ^{ - 1}}\left( {{{ - {B \over 2}} \over {A - B{{\sqrt 3 } \over 2}}}} \right)$$
B
$${\tan ^{ - 1}}\left( {{A \over {0.7B}}} \right)$$
C
$${\tan ^{ - 1}}\left( {{{\sqrt 3 B} \over {2A - B}}} \right)$$
D
$${\tan ^{ - 1}}\left( {{{B\cos \theta } \over {A - B\sin \theta }}} \right)$$

Explanation



Angle between $${\overrightarrow A }$$ and $${\overrightarrow B }$$, $$\theta$$ = 60$$^\circ$$

Angle between $${\overrightarrow A }$$ and -$${\overrightarrow B }$$, $$\theta$$ = 120$$^\circ$$

If angle between $${\overrightarrow A }$$ and $${\overrightarrow A }$$ $$-$$ $${\overrightarrow B }$$ is $$\alpha $$

then $$\tan \alpha = $$$${{\left| { - \overrightarrow B } \right|\sin \theta } \over {\overrightarrow A + \left| { - \overrightarrow B } \right|\cos \theta }}$$

= $${{B\sin 120^\circ } \over {A + B\cos 120^\circ }}$$

=$${{B{{\sqrt 3 } \over 2}} \over {A - {B \over 2}}}$$

$$ \Rightarrow $$ $$\tan \alpha = {{\sqrt 3 B} \over {2A - B}}$$
4

JEE Main 2021 (Online) 26th August Morning Shift

MCQ (Single Correct Answer)
The magnitude of vectors $$\overrightarrow {OA} $$, $$\overrightarrow {OB} $$ and $$\overrightarrow {OC} $$ in the given figure are equal. The direction of $$\overrightarrow {OA} $$ + $$\overrightarrow {OB} $$ $$-$$ $$\overrightarrow {OC} $$ with x-axis will be :

A
$${\tan ^{ - 1}}{{(1 - \sqrt 3 - \sqrt 2 )} \over {(1 + \sqrt 3 + \sqrt 2 )}}$$
B
$${\tan ^{ - 1}}{{(\sqrt 3 - 1 + \sqrt 2 )} \over {(1 + \sqrt 3 - \sqrt 2 )}}$$
C
$${\tan ^{ - 1}}{{(\sqrt 3 - 1 + \sqrt 2 )} \over {(1 - \sqrt 3 + \sqrt 2 )}}$$
D
$${\tan ^{ - 1}}{{(1 + \sqrt 3 - \sqrt 2 )} \over {(1 - \sqrt 3 - \sqrt 2 )}}$$

Explanation


Let magnitude be equal to $$\lambda$$

$$\overrightarrow {OA} = \lambda \left[ {\cos 30^\circ \widehat i + \sin 30\widehat j} \right] = \lambda \left[ {{{\sqrt 3 } \over 2}\widehat i + {1 \over 2}\widehat j} \right]$$

$$\overrightarrow {OB} = \lambda \left[ {\cos 60^\circ \widehat i - \sin 60\widehat j} \right] = \lambda \left[ {{1 \over 2}\widehat i - {{\sqrt 3 } \over 2}\widehat j} \right]$$

$$\overrightarrow {OC} = \lambda \left[ {\cos 45^\circ ( - \widehat i) + \sin 45\widehat j} \right] = \lambda \left[ { - {1 \over {\sqrt 2 }}\widehat i + {1 \over {\sqrt 2 }}\widehat j} \right]$$

$$\therefore$$ $$\overrightarrow {OA} + \overrightarrow {OB} - \overrightarrow {OC} $$

$$ = \lambda \left[ {\left( {{{\sqrt 3 + 1} \over 2} + {1 \over {\sqrt 2 }}} \right)\widehat i + \left( {{1 \over 2} - {{\sqrt 3 } \over 2} - {1 \over {\sqrt 2 }}} \right)\widehat j} \right]$$

$$\therefore$$ Angle with x-axis

$${\tan ^{ - 1}}\left[ {{{{1 \over 2} - {{\sqrt 3 } \over 2} - {1 \over {\sqrt 2 }}} \over {{{\sqrt 3 } \over 2} + {1 \over 2} + {1 \over {\sqrt 2 }}}}} \right] = {\tan ^{ - 1}}\left[ {{{\sqrt 2 - \sqrt 6 - 2} \over {\sqrt 6 + \sqrt 2 + 2}}} \right]$$

$$ = {\tan ^{ - 1}}\left[ {{{1 - \sqrt 3 - \sqrt 2 } \over {\sqrt 3 + 1 + \sqrt 2 }}} \right]$$

Hence option (a).

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