1
JEE Main 2018 (Online) 15th April Evening Slot
+4
-1
When an air bubble of radius r rises from the bottom to the surface of a lake its radius becomes $${{5r} \over 4}.$$ Taking the atmospheric pressure to be equal to 10 m height of water column, the depth of the lake would approximately be (ignore the surface tension and the effect of temperature) :
A
11.2 m
B
8.7 m
C
9.5 m
D
10.5 m
2
JEE Main 2018 (Online) 15th April Evening Slot
+4
-1
As shown in the figure, forces of 105 N each are applied in opposite directions, on the upper and lower faces of a cube of side 10 cm, shifting the upper face parallel to itself by 0.5 cm. If the side of another cube of the same material is 20 cm, then under similar conditions as above, the displacement will be :
A
0.25 cm
B
0.37 cm
C
0.75 cm
D
1.00 cm
3
JEE Main 2018 (Online) 15th April Evening Slot
+4
-1
Out of Syllabus
A body takes 10 minutes to cool from 60oC to 50oC. The tempertature of surroundings is constant at 25oC. Then, the temperature of the body after next 10 minutes will be approximately :
A
47oC
B
41oC
C
45oC
D
43oC
4
JEE Main 2018 (Online) 15th April Morning Slot
+4
-1
A thin uniform tube is bent into a circle of radius $$r$$ in the vertical plane. Equal volumes of two immiscible liquids, whose densities are $${\rho _1}$$ and $${\rho _2}$$ $$\left( {{\rho _1} > {\rho _2}} \right),$$ fill half the circle. The angle $$\theta$$ between the radius vector passing through the common interface and the vertical is :
A
$$\theta = {\tan ^{ - 1}}\pi \left( {{{{\rho _1}} \over {{\rho _2}}}} \right)$$
B
$$\theta = {\tan ^{ - 1}}{\pi \over 2}\left( {{{{\rho _1}} \over {{\rho _2}}}} \right)$$
C
$$\theta = {\tan ^{ - 1}}\left( {{{{\rho _1} - {\rho _2}} \over {{\rho _1} + {\rho _2}}}} \right)$$
D
$$\theta = {\tan ^{ - 1}}{\pi \over 2}\left( {{{{\rho _1} + {\rho _2}} \over {{\rho _1} - {\rho _2}}}} \right)$$
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