BITSAT 2026 May 26 Shift 1 Question Paper: Download Answer Key with Solutions PDF

A particle moves along a circle of radius \(R\) with a constant angular acceleration \(\alpha\). If the initial angular velocity is zero, the total acceleration of the particle at time \(t\) is:

• (A) \( R\alpha \)
• (B) \( R\alpha^2 t^2 \)
• (C) \( R\alpha\sqrt{1 + \alpha^2 t^4} \)
• (D) \( R\alpha t \)

The focal length of a convex lens is \(f\) in air. When it is completely immersed in water of refractive index \(\frac{4}{3}\), its focal length becomes (take refractive index of glass = 1.5):

• (A) \( f \)
• (B) \( 2f \)
• (C) \( 4f \)
• (D) \( \frac{f}{2} \)

The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength \(\lambda\) is \(V_s\). If the intensity of the incident light is doubled while keeping wavelength identical, the stopping potential will be:

• (A) \( 2V_s \)
• (B) \( \frac{V_s}{2} \)
• (C) \( V_s \)
• (D) \( 4V_s \)

The de-Broglie wavelength of an electron accelerated from rest through a potential difference of \(100V\) is approximately:

• (A) \( 1.227\,\text{\AA} \)
• (B) \( 12.27\,\text{\AA} \)
• (C) \( 0.1227\,\text{\AA} \)
• (D) \( 122.7\,\text{\AA} \)

Balance the following redox reaction in acidic medium and determine the stoichiometric coefficient of \(H_2O\) in the final balanced equation. \[ MnO_4^-(aq) + Fe^{2+}(aq) \rightarrow Mn^{2+}(aq) + Fe^{3+}(aq) \]

• (A) 2
• (B) 4
• (C) 6
• (D) 8

Titration of \(0.1467 g\) of primary standard \(Na_2C_2O_4\) required \(28.85 mL\) of \(KMnO_4\) solution. Calculate the molar concentration of \(KMnO_4\) solution.

• (A) \( 0.01518 M \)
• (B) \( 0.001518 M \)
• (C) \( 0.15180 M \)
• (D) \( 1.5180 M \)

A current of \(4.0 A\) is passed through \(0.5 L\) of \(0.2 M NaCl\) solution for \(1200s\). Calculate the \(pH\) of the solution after electrolysis.

• (A) \( 1.3 \)
• (B) \( 13 \)
• (C) \( 7.0 \)
• (D) \( 2.0 \)

Using the standard electrode potential, find out the pair between which redox reaction is not feasible.

\[ \begin{aligned} E^\ominus \text{ values: } & \\ Fe^{3+}/Fe^{2+} &= +0.77\,V \\ I_2/I^- &= +0.54\,V \\ Cu^{2+}/Cu &= +0.34\,V \\ Ag^+/Ag &= +0.80\,V \end{aligned} \]

• (A) \( Fe^{3+} \text{ and } I^- \)
• (B) \( Ag^+ \text{ and } Cu \)
• (C) \( Fe^{3+} \text{ and } Cu \)
• (D) \( Ag \text{ and } Fe^{3+} \)

If \(p\) and \(q\) be the longest and the shortest distance respectively of the point \((-7, 2)\) from any point \((\alpha, \beta)\) on the curve whose equation is \(x^2 + y^2 - 10x - 14y - 51 = 0\), then find the Geometric Mean (G.M.) of \(p\) and \(q\).

• (A) \( 2\sqrt{11} \)
• (B) \( 5\sqrt{5} \)
• (C) \( 13 \)
• (D) \( 11 \)

The distance from the origin to the image of \((1, 1)\) with respect to the line \(x + y + 5 = 0\) is:

• (A) \( 7\sqrt{2} \)
• (B) \( 3\sqrt{2} \)
• (C) \( 6\sqrt{2} \)
• (D) \( 4\sqrt{2} \)

General solution of \(\tan 5\theta = \cot 2\theta\) is:

• (A) \( \theta = \frac{n\pi}{7} + \frac{\pi}{14} \)
• (B) \( \theta = \frac{n\pi}{7} + \frac{\pi}{5} \)
• (C) \( \theta = \frac{n\pi}{7} + \frac{\pi}{2} \)
• (D) \( \theta = \frac{n\pi}{7} + \frac{\pi}{3} \)

The sum of the series \(\left(x + \frac{1}{x}\right)^2 + \left(x^2 + \frac{1}{x^2}\right)^2 + \left(x^3 + \frac{1}{x^3}\right)^2 \dots\dots\dots\) up to \(n\) terms is:

• (1) \( \frac{x^{2n} - 1}{x^2 - 1} \times \frac{x^{2n+2} + 1}{x^{2n}} + 2n \)
• (2) \( \frac{x^{2n} + 1}{x^2 + 1} \times \frac{x^{2n+2} - 1}{x^{2n}} - 2n \)
• (3) \( \frac{x^{2n} - 1}{x^2 - 1} \times \frac{x^{2n} - 1}{x^{2n}} - 2n \)
• (4) None of these