1
JEE Main 2019 (Online) 10th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
If y = y(x) is the solution of the differential equation
$${{dy} \over {dx}} = \left( {\tan x - y} \right){\sec ^2}x$$, $$x \in \left( { - {\pi \over 2},{\pi \over 2}} \right)$$,
such that y (0) = 0, then $$y\left( { - {\pi \over 4}} \right)$$ is equal to :
A
$${1 \over 2} - e$$
B
$$e - 2$$
C
$$2 + {1 \over e}$$
D
$${1 \over e} - 2$$
2
JEE Main 2019 (Online) 10th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
All the pairs (x, y) that satisfy the inequality
$${2^{\sqrt {{{\sin }^2}x - 2\sin x + 5} }}.{1 \over {{4^{{{\sin }^2}y}}}} \le 1$$
also satisfy the equation
A
sin x = |sin y|
B
sin x = 2sin y
C
2 sin x = sin y
D
2 |sin x | = 3 sin y
3
JEE Main 2019 (Online) 10th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
The ratio of surface tensions of mercury and water is given to be 7.5 while the ratio of thier densities is 13.6. Their contact angles, with glass, are close to 135° and 0°, respectively. It is observed that mercury gets depressed by an amount h in a capillary tube of radius r1, while water rises by the same amount h in a capillary tube of radius r2. The ratio, (r1/r2), is then close to :
A
2/5
B
2/3
C
3/5
D
4/5
4
JEE Main 2019 (Online) 10th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Two wires A & B are carrying currents I1 & I2 as shown in the figure. The separation between them is d. A third wire C carrying a current I is to be kept parallel to them at a distance x from A such that the net force acting on it is zero. The possible values of x are : JEE Main 2019 (Online) 10th April Morning Slot Physics - Magnetic Effect of Current Question 179 English
A
$$x = \left( {{{{I_1}} \over {{I_1} + {I_2}}}} \right)d\,and\,x = \left( {{{{I_2}} \over {{I_1} - {I_2}}}} \right)d$$
B
$$x = \left( {{{{I_2}} \over {{I_1} + {I_2}}}} \right)d\,and\,x = \left( {{{{I_2}} \over {{I_1} - {I_2}}}} \right)d$$
C
$$x = \left( {{{{I_1}} \over {{I_1} - {I_2}}}} \right)d\,and\,x = \left( {{{{I_2}} \over {{I_1} + {I_2}}}} \right)d$$
D
$$x = \pm {{{I_1}d} \over {{I_1} - {I_2}}}$$

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