1
JEE Main 2024 (Online) 8th April Morning Shift
MCQ (Single Correct Answer)
+4
-1

Young's modulus is determined by the equation given by $$\mathrm{Y}=49000 \frac{\mathrm{m}}{\mathrm{l}} \frac{\mathrm{dyne}}{\mathrm{cm}^2}$$ where $$M$$ is the mass and $$l$$ is the extension of wire used in the experiment. Now error in Young modules $$(Y)$$ is estimated by taking data from $$M-l$$ plot in graph paper. The smallest scale divisions are $$5 \mathrm{~g}$$ and $$0.02 \mathrm{~cm}$$ along load axis and extension axis respectively. If the value of $M$ and $l$ are $$500 \mathrm{~g}$$ and $$2 \mathrm{~cm}$$ respectively then percentage error of $$Y$$ is :

A
2%
B
0.02%
C
0.5%
D
0.2%
2
JEE Main 2024 (Online) 8th April Morning Shift
MCQ (Single Correct Answer)
+4
-1

Correct Bernoulli's equation is (symbols have their usual meaning) :

A
$$P+\frac{1}{2} \rho g h+\frac{1}{2} \rho v^2=$$ constant
B
$$P+m g h+\frac{1}{2} m v^2=$$ constant
C
$$P+\rho g h+\rho v^2=$$ constant
D
$$P+\rho g h+\frac{1}{2} \rho v^2=$$ constant
3
JEE Main 2024 (Online) 6th April Evening Shift
MCQ (Single Correct Answer)
+4
-1

Pressure inside a soap bubble is greater than the pressure outside by an amount : (given : $$\mathrm{R}=$$ Radius of bubble, $$\mathrm{S}=$$ Surface tension of bubble)

A
$$\frac{S}{R}$$
B
$$\frac{4 \mathrm{~S}}{\mathrm{R}}$$
C
$$\frac{4 \mathrm{R}}{\mathrm{S}}$$
D
$$\frac{2 S}{R}$$
4
JEE Main 2024 (Online) 6th April Morning Shift
MCQ (Single Correct Answer)
+4
-1

A small ball of mass $$m$$ and density $$\rho$$ is dropped in a viscous liquid of density $$\rho_0$$. After sometime, the ball falls with constant velocity. The viscous force on the ball is :

A
$$m g\left(1-\frac{\rho_0}{\rho}\right)$$
B
$$m g\left(\frac{\rho_0}{\rho}-1\right)$$
C
$$m g\left(1-\rho \rho_0\right)$$
D
$$m g\left(1+\frac{\rho}{\rho_0}\right)$$
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