1
JEE Main 2026 (Online) 5th April Morning Shift
Numerical
+4
-1
Change Language

Let $y=y(x)$ be the solution of the differential equation $x \sin \left(\frac{y}{x}\right) d y=\left(y \sin \left(\frac{y}{x}\right)-x\right) d x, y(1)=\frac{\pi}{2}$ and let $\alpha=\cos \left(\frac{y\left(e^{12}\right)}{e^{12}}\right)$. Then the number of integral value of $p$, for which the equation $x^2+y^2-2 p x+2 p y+\alpha+2=0$ represents a circle of radius $r \leq 6$, is $\_\_\_\_$ .

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2
JEE Main 2026 (Online) 5th April Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

In a Vernier calipers, when both jaws touch each other, zero of the Vernier scale is shifted to the right of zero of the main scale and $7^{\text {th }}$ Vernier division coincides with a main scale reading. If the value of 1 main scale division is 1 mm and there are 10 Vernier scale divisions, then the Vernier caliper has

A

0.07 cm negative zero error

B

0.7 cm negative zero error

C

0.07 cm positive zero error

D

0.7 cm positive zero error

3
JEE Main 2026 (Online) 5th April Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

$L, C$ and $R$ represents physical quantities inductance, capacitance and resistance respectively. The dimensional formula $\mathrm{ML}^2 \mathrm{~T}^{-4} \mathrm{~A}^{-2}$ corresponds to $\_\_\_\_$ .

A

$\frac{R}{\sqrt{L C}}$

B

$\frac{R}{L C}$

C

$\frac{C}{\sqrt{L R}}$

D

$\frac{1}{R} \sqrt{\frac{L}{C}}$

4
JEE Main 2026 (Online) 5th April Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

When one moves from a point 16 km below the earth's surface to a point 16 km above the earth's surface. The change in g is approximately $\alpha \%$. The value of $\alpha$ is $\_\_\_\_$ .

(Take radius of the earth $=6400 \mathrm{~km}$.)

A

0.12

B

0.25

C

0.50

D

0.75

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