1
JEE Main 2022 (Online) 28th June Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Velocity (v) and acceleration (a) in two systems of units 1 and 2 are related as $${v_2} = {n \over {{m^2}}}{v_1}$$ and $${a_2} = {{{a_1}} \over {mn}}$$ respectively. Here m and n are constants. The relations for distance and time in two systems respectively are :

A
$${{{n^3}} \over {{m^3}}}{L_1} = {L_2}$$ and $${{{n^2}} \over m}{T_1} = {T_2}$$
B
$${L_1} = {{{n^4}} \over {{m^2}}}{L_2}$$ and $${T_1} = {{{n^2}} \over m}{T_2}$$
C
$${L_1} = {{{n^2}} \over m}{L_2}$$ and $${T_1} = {{{n^4}} \over {{m^2}}}{T_2}$$
D
$${{{n^2}} \over m}{L_1} = {L_2}$$ and $${{{n^4}} \over {{m^2}}}{T_1} = {T_2}$$
2
JEE Main 2022 (Online) 27th June Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

A projectile is launched at an angle '$$\alpha$$' with the horizontal with a velocity 20 ms$$-$$1. After 10 s, its inclination with horizontal is '$$\beta$$'. The value of tan$$\beta$$ will be : (g = 10 ms$$-$$2).

A
tan$$\alpha$$ + 5sec$$\alpha$$
B
tan$$\alpha$$ $$-$$ 5sec$$\alpha$$
C
2tan$$\alpha$$ $$-$$ 5sec$$\alpha$$
D
2tan$$\alpha$$ $$+$$ 5sec$$\alpha$$
3
JEE Main 2022 (Online) 27th June Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

A girl standing on road holds her umbrella at 45$$^\circ$$ with the vertical to keep the rain away. If she starts running without umbrella with a speed of 15$$\sqrt2$$ kmh$$-$$1, the rain drops hit her head vertically. The speed of rain drops with respect to the moving girl is :

A
30 kmh$$-$$1
B
$${{25} \over {\sqrt 2 }}$$ kmh$$-$$1
C
$${{30} \over {\sqrt 2 }}$$ kmh$$-$$1
D
25 kmh$$-$$1
4
JEE Main 2022 (Online) 25th June Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Given below are two statements. One is labelled as Assertion A and the other is labelled as Reason R.

Assertion A : Two identical balls A and B thrown with same velocity 'u' at two different angles with horizontal attained the same range R. IF A and B reached the maximum height h1 and h2 respectively, then $$R = 4\sqrt {{h_1}{h_2}} $$

Reason R : Product of said heights.

$${h_1}{h_2} = \left( {{{{u^2}{{\sin }^2}\theta } \over {2g}}} \right)\,.\,\left( {{{{u^2}{{\cos }^2}\theta } \over {2g}}} \right)$$

Choose the correct answer :

A
Both A and R are true and R is the correct explanation of A.
B
Both A and R are true but R is NOT the correct explanation of A.
C
A is true but R is false.
D
A is false but R is true.
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