Mathematics
1. Evaluate the expression: $\csc 10^\circ – \sqrt{3} \sec 10^\circ$
a) 8
b) 6
c) 4
d) 2
(JEE Main 2026 | Exam Date: 21.01.2026 | Shift I)
Answer: (c) 4
Solution: Convert the terms into sine and cosine:
$$\frac{1}{\sin 10^\circ} – \frac{\sqrt{3}}{\cos 10^\circ} = \frac{\cos 10^\circ – \sqrt{3} \sin 10^\circ}{\sin 10^\circ \cos 10^\circ}$$
Multiply and divide the numerator by 2:
$$\frac{2\left(\frac{1}{2}\cos 10^\circ – \frac{\sqrt{3}}{2}\sin 10^\circ\right)}{\sin 10^\circ \cos 10^\circ} = \frac{2(\sin 30^\circ \cos 10^\circ – \cos 30^\circ \sin 10^\circ)}{\sin 10^\circ \cos 10^\circ}$$
Apply the identity $\sin(A-B)$:
$$\frac{2\sin(30^\circ – 10^\circ)}{\sin 10^\circ \cos 10^\circ} = \frac{2\sin 20^\circ}{\frac{1}{2}\sin 20^\circ} = 4$$
2. The sum of all roots of the equation $(x – 1)^2 – 5|x – 1| + 6 = 0$ is:
a) 4
b) 1
c) 5
d) 3
(JEE Main 2026 | Exam Date: 21.01.2026 | Shift I)
Answer: (a) 4
Solution: Let $|x – 1| = t$. The equation becomes $t^2 – 5t + 6 = 0$. Solving for $t$:
$$(t – 2)(t – 3) = 0 \Rightarrow t = 2, 3$$
If $|x – 1| = 2$, then $x – 1 = 2$ or $x – 1 = -2$, giving $x = 3, -1$.
If $|x – 1| = 3$, then $x – 1 = 3$ or $x – 1 = -3$, giving $x = 4, -2$.
Sum of roots: $3 + (-1) + 4 + (-2) = 4$.
Physics
1. A rod of mass $m$ and length $l$ is attached to two ideal strings. Find tension in left string just after right string is cut.
(1) $\frac{mg}{2}$
(2) $\frac{mg}{4}$
(3) $\frac{2mg}{3}$
(4) $\frac{mg}{5}$
Answer: (2)
Solution:
Mass of rod is \( m \) and length \( l \).Step 1: Before the right string is cut
The rod is in equilibrium.\[
T_L + T_R = mg
\]Taking moments about the center of mass (midpoint of the rod):
\[
T_R \cdot \frac{l}{2} = T_L \cdot \frac{l}{2}
\]
Hence,
\[
T_L = T_R = \frac{mg}{2}
\]
Step 2: Just after the right string is cut
The right tension \( T_R = 0 \).
The left tension \( T_L’ \) acts upward at the left end.
The weight \( mg \) acts downward at the center of mass.
The rod will start rotating about the left end.
Step 3: Equations of motion
Let the angular acceleration of the rod about the left end be \( \alpha \).
Taking moments about the left end:
\[
mg \cdot \frac{l}{2} = I_{\text{left}} \cdot \alpha
\]
Moment of inertia of a uniform rod about one end:
\[
I_{\text{left}} = \frac{1}{3} m l^2
\]
Substituting:
\[
mg \cdot \frac{l}{2} = \frac{1}{3} m l^2 \alpha
\]
\[
\alpha = \frac{3g}{2l}
\]
Step 4: Linear acceleration of the center of mass
The center of mass is at a distance \( \frac{l}{2} \) from the left end.
\[
a_{\text{CM}} = \alpha \cdot \frac{l}{2}
\]
\[
a_{\text{CM}} = \frac{3g}{2l} \cdot \frac{l}{2} = \frac{3g}{4}
\]
Step 5: Vertical force balance for the whole rod
Taking upward as positive:
\[
T_L’ – mg = m a_{\text{CM, vertical}}
\]
Since the acceleration is downward,
\[
a_{\text{CM, vertical}} = -\frac{3g}{4}
\]
\[
T_L’ – mg = -m \frac{3g}{4}
\]
\[
T_L’ = mg – \frac{3mg}{4} = \frac{mg}{4}
\]
2. Which logic gate is given in the figure?
(1) XOR
(2) NOR
(3) NAND
(4) OR
Answer: (3)
Solution:
- The inputs $A$ and $B$ are each passed through a NOT gate, resulting in $\bar{A}$ and $\bar{B}$.
- These inverted signals are then fed into a NOR gate. The output of the NOR gate is:$\text{Output}_{\text{NOR}} = \overline{\bar{A} + \bar{B}}$
- Applying De Morgan’s Law, this expression simplifies:$\overline{\bar{A} + \bar{B}} = \overline{\bar{A}} \cdot \overline{\bar{B}} = A \cdot B$
- Finally, this signal $A \cdot B$ passes through another NOT gate before reaching the output $C$:$C = \overline{A \cdot B}$
3. Find dimensions of $A/B$ if $\rho \left( \frac{At^2}{B} + V^2 + \frac{P}{\rho} \right) = \text{constant}$ where $P \to \text{pressure}, \rho \to \text{density}, V \to \text{speed}$.
(1) $ML^{1}T^{-4}$
(2) $ML^{-1}T^{-4}$
(3) $ML^{2}T^{-4}$
(4) $ML^{-1}T^{-2}$
Answer: (2)
Solution:
According to the principle of homogeneity, only physical quantities with the same dimensions can be added or subtracted. In the given expression, the terms inside the parentheses must share the same dimensional formula.
The solution provided in the sources establishes that the term $\frac{At^2}{B}$ shares the dimensions of pressure ($P$):
- Identify dimensions of Pressure ($P$):$[P] = [ML^{-1}T^{-2}]$
- Equate the dimensions:$[P] = \left[ \frac{At^2}{B} \right]$
- Substitute known dimensions:$[ML^{-1}T^{-2}] = \left[ \frac{A}{B} \right] [T^2]$
- Solve for the ratio $A/B$:$\left[ \frac{A}{B} \right] = \frac{ML^{-1}T^{-2}}{T^2}$$\left[ \frac{A}{B} \right] = \mathbf{ML^{-1}T^{-4}}$
4. Find the heat produced in external circuit (AB) in one minute.
(1) 1181.25 J
(2) 1311.25 J
(3) 1207.50 J
(4) 1410.50 J
Answer: (1)
Solution:
Using Kirchhoff’s law or star-delta transformation for the circuit provided in the diagram:
Equivalent resistance of external circuit $R_{AB} = 1.4\ \Omega$
Total current $i = 3.75\ \text{A}$
Power $P = i^2 R_{AB}$
Heat $H = P \times t = 1181.25\ \text{J}$