BITSAT 2026 May 25 Shift 1 Question Paper: Download Answer Key with Solutions PDF

Concept:
Electric potential (\(V\)) is a scalar quantity, so the total potential at a point is the simple algebraic sum of the potentials due to individual charges. \[ V_{net} = \sum V_i = \frac{k q_i}{r_i} \]
Electric field (\(E\)) is a vector quantity, so the net electric field is the vector sum of individual fields, taking both magnitude and direction into account. \[ \vec{E}_{net} = \vec{E}_1 + \vec{E}_2 + \dots \]


Step 1: Calculating the net electric potential (\(V\)) at the midway point.

Let the distance of the midway point from both charges be \( r = \frac{d}{2} \). \[ V_{net} = V_+ + V_- = \frac{k q}{\frac{d}{2}} + \frac{k (-q)}{\frac{d}{2}} = \frac{2kq}{d} - \frac{2kq}{d} = 0 \]


Step 2: Calculating the net electric field (\(E\)) at the midway point.

The electric field due to the positive charge (\(+q\)) points away from it (towards the right). The electric field due to the negative charge (\(-q\)) points towards it (also towards the right).

Since both field vectors point in the same direction, their magnitudes add up: \[ E_{net} = E_+ + E_- = \frac{k q}{\left(\frac{d}{2}\right)^2} + \frac{k q}{\left(\frac{d}{2}\right)^2} = \frac{4kq}{d^2} + \frac{4kq}{d^2} = \frac{8kq}{d^2} \neq 0 \] Quick Tip: For a dipole system, at any point on the perpendicular bisector (including the midpoint), the net electric potential is always zero, while the net electric field is non-zero and parallel to the dipole axis.

Concept:
Inside a conductor under electrostatic equilibrium, the electric field is zero everywhere. When a charge is placed inside the cavity of a conducting shell, electrostatic induction occurs:

An equal and opposite charge is induced on the inner surface of the shell.
An equal and similar charge is induced on the outer surface of the shell to maintain its initial net charge state.

The surface charge density (\(\sigma\)) is given by the formula: \[ \sigma = \frac{Charge on the surface}{Surface area} = \frac{q}{4\pi R^2} \]


Step 1: Determining the induced charges on the inner and outer surfaces.

The initial net charge on the conducting spherical shell is zero (uncharged). When a charge \(+Q\) is placed at its center: \[ Induced charge on the inner surface (at radius R_1) = -Q \]
To keep the net charge of the shell zero, a balancing positive charge must appear on its exterior boundary: \[ Induced charge on the outer surface (at radius R_2) = +Q \]


Step 2: Calculating the surface charge density on the outer surface.

The surface area of the outer sphere with radius \(R_2\) is \(4\pi R_2^2\). Substituting the outer charge and surface area into the density formula gives: \[ \sigma_{outer} = \frac{Charge on outer surface}{Outer surface area} = \frac{Q}{4\pi R_2^2} \] Quick Tip: By Gauss's Law, the total charge enclosed within any Gaussian surface drawn inside the conducting material must be zero, forcing the inner surface charge to be completely opposite to the central cavity charge.

Concept:
When a capacitor is disconnected from its charging battery, the isolated system can no longer exchange charge with any source. Therefore, the charge \(Q\) remains constant.
The capacitance \(C\), potential difference \(V\), and electrostatic energy stored \(U\) depend on the geometric configuration formulas:

Capacitance: \( C = \frac{\varepsilon_0 A}{d} \)
Potential Difference: \( V = \frac{Q}{C} \)
Electrostatic Energy: \( U = \frac{Q^2}{2C} \)



Step 1: Analyzing the effect of pulling the plates further apart.

When the plates are pulled further apart, the separation distance \(d\) between them increases.
From the capacitance relation: \[ C \propto \frac{1}{d} \quad \Rightarrow \quad As d increases, C decreases. \]
Hence, Option (C) is incorrect.


Step 2: Evaluating the change in voltage and stored energy.

Since the capacitor is disconnected from the battery, the charge \(Q\) must stay constant (\(Q = constant\)). Hence, Option (A) is incorrect.
Using the potential difference relationship: \[ V = \frac{Q}{C} \quad \Rightarrow \quad Since C decreases, V must increase. \]
Hence, Option (B) is incorrect.

Now, evaluating the stored electrostatic energy \(U\): \[ U = \frac{Q^2}{2C} \quad \Rightarrow \quad Since Q is constant and C decreases, U must increase. \]
This additional stored energy comes directly from the mechanical work done by an external agent against the attractive electrostatic forces between the oppositely charged plates. Quick Tip: Remember the golden rule for capacitor transformations: Battery disconnected \(\Rightarrow Q\) remains constant. Battery remains connected \(\Rightarrow V\) remains constant.

Concept:
The magnetic field due to a circular loop carrying a current \(I\) depends heavily on the evaluation point.

At the center of the loop, the magnetic field is given by:
\[ B_{center} = \frac{\mu_0 I}{2R} \]
At any axial point at a distance \(x\) from the center, the magnetic field formula is:
\[ B_{axial} = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}} \]



Step 1: Expressing the given magnetic field at the center.

We are given that the magnetic field at the center is \(B_0\). Therefore: \[ B_0 = \frac{\mu_0 I}{2R} \quad \cdots (1) \]


Step 2: Calculating the magnetic field at the axial distance \(x = R\).

Substitute \(x = R\) into the axial magnetic field formula: \[ B_{axial} = \frac{\mu_0 I R^2}{2(R^2 + R^2)^{3/2}} = \frac{\mu_0 I R^2}{2(2R^2)^{3/2}} \]
Simplifying the denominator \((2R^2)^{3/2}\): \[ (2R^2)^{3/2} = 2^{3/2} \cdot (R^2)^{3/2} = 2\sqrt{2} R^3 \]
Now, substitute this simplified expression back into the equation: \[ B_{axial} = \frac{\mu_0 I R^2}{2 \cdot 2\sqrt{2} R^3} = \frac{\mu_0 I}{2R \cdot 2\sqrt{2}} \]
Using equation (1), replace \(\frac{\mu_0 I}{2R}\) with \(B_0\): \[ B_{axial} = \frac{B_0}{2\sqrt{2}} \] Quick Tip: For rapid calculations on a circular loop's axis, remember the ratio formula: \( \frac{B_{axial}}{B_{center}} = \left(1 + \frac{x^2}{R^2}\right)^{-3/2} \). Substituting \(x = R\) directly yields \((1+1)^{-3/2} = 2^{-3/2} = \frac{1}{2\sqrt{2}}\).

Concept:
The relationship between enthalpy change (\(\Delta H\)) and internal energy change (\(\Delta U\)) for a chemical reaction at a constant temperature \(T\) is given by the formula: \[ \Delta H = \Delta U + \Delta n_g RT \]
Where:

\(\Delta n_g\) = (Sum of stoichiometric coefficients of gaseous products) \(-\) (Sum of stoichiometric coefficients of gaseous reactants)
\(R\) = Universal gas constant = \(8.314 J mol^{-1} K^{-1} = 8.314 \times 10^{-3} kJ mol^{-1} K^{-1}\)
\(T\) = Absolute temperature in Kelvin



Step 1: Calculating the change in gaseous moles (\(\Delta n_g\)) and converting temperature to Kelvin.

Look at the physical states given in the balanced chemical equation: \[ C_2H_5OH(l) + 3O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l) \]

Gaseous products: \(2 moles of CO_2(g)\)
Gaseous reactants: \(3 moles of O_2(g)\)
\[ \Delta n_g = 2 - 3 = -1 \]

Convert temperature to Kelvin:

Concept:
A bomb calorimeter operates at a constant volume. Therefore, the heat produced or exchanged in a bomb calorimeter is equal to the internal energy change (\(\Delta U\)) of the reaction. Because heat is *produced* (exothermic combustion), \(\Delta U\) is negative.
The relation between the enthalpy of combustion (\(\Delta H_C\)) and the internal energy change (\(\Delta U\)) is given by: \[ \Delta H_C = \Delta U + \Delta n_g RT \]
Where:

\(\Delta n_g\) = (Sum of moles of gaseous products) \(-\) (Sum of moles of gaseous reactants)
\(R\) = Universal gas constant = \(8.314 \times 10^{-3} kJ mol^{-1} K^{-1}\)
\(T\) = Absolute temperature in Kelvin



Step 1: Determining \(\Delta U\), \(\Delta n_g\), and converting temperature to Kelvin.

Given that the heat produced at constant volume is \(1364.47 kJ mol^{-1}\): \[ \Delta U = -1364.47 kJ mol^{-1} \]
From the balanced chemical equation: \[ C_2H_5OH(l) + 3O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l) \]
The gaseous components are \(2 moles of CO_2(g)\) and \(3 moles of O_2(g)\). Liquids are omitted. \[ \Delta n_g = 2 - 3 = -1 \]

Convert the given temperature to Kelvin:

Concept:
This is a problem based on the common ion effect.

Potassium hydroxide (\(KOH\)) is a strong base and dissociates completely in water to yield a high concentration of hydroxyl ions (\(OH^-\)).
Ammonia (\(NH_3\)) is a weak base that partially ionizes in water to form \(NH_4^+\) and \(OH^-\).
The presence of the common ion \(OH^-\) from the strong base shifts the equilibrium of the weak base backward, suppressing its dissociation even further.

Step 1: Determine the equilibrium concentrations.

First, write the complete dissociation of the strong base:

Since \([KOH] = 0.01\,M\), the concentration of \(OH^-\) contributed by \(KOH\) is:

Now write the equilibrium for ammonia:

ICE table:

Since \(K_b\) is very small (\(1.8\times10^{-5}\)) and dissociation is highly suppressed because of the common ion effect:

Step 2: Calculate \([NH_4^+]\).

Using the base dissociation constant expression:

Substitute the equilibrium values:

Simplify:

Therefore,

Quick Tip: When a weak base is mixed with a strong base, you can directly use the shortcut:

Here:

Concept:
The equilibrium constant (\(K_c\)) for a generalized reversible chemical reaction is expressed as the ratio of the product of the equilibrium molar concentrations of the products to that of the reactants, with each concentration term raised to a power equal to its stoichiometric coefficient in the balanced chemical equation.

The equilibrium constant expression is:

Where

Step 1: Setting up the equilibrium table in terms of degree of dissociation (\(x\)).

Let \(2x\) moles of \(C\) be formed at equilibrium based on its stoichiometric coefficient. The ICE (Initial, Change, Equilibrium) setup for the moles is as follows:

Given that the number of moles of \(C\) at equilibrium is \(0.2\):

Step 2: Calculating equilibrium concentrations and computing \(K_c\).

Substitute the value of \(x=0.1\) into the equilibrium expressions:

Moles of \(A = 1.1 - 0.1 = 1.0\) mole
Moles of \(B = 2.2 - 2(0.1) = 2.0\) moles
Moles of \(C = 0.2\) mole
Moles of \(D = 0.1\) mole

Since the volume of the container is \(1\,L\), molarity equals the number of moles:

Substitute these values into the equilibrium constant expression:

Quick Tip: Always check the container volume first. When the volume is \(1\,L\), moles and molarities become numerically equal, which simplifies equilibrium calculations significantly.

Concept:
When a variable exists in the exponent of a function, we apply **logarithmic differentiation**. Taking the natural logarithm (\(\log_e\) or \(\ln\)) on both sides simplifies the exponents into linear products using the logarithm power rule: \[ \log(a^b) = b \log a \]
Once simplified, we can express \(x\) explicitly in terms of \(y\) to compute \(\frac{dx}{dy}\), and then find the derivative using the reciprocal identity: \[ \frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} \]


Step 1: Taking the natural logarithm on both sides and simplifying.

Given equation: \[ y^x = e^{y - x} \]
Taking natural log (\(\log\)) on both sides: \[ \log(y^x) = \log(e^{y - x}) \] \[ x \log y = (y - x) \log e \]
Since \(\log e = 1\), the expression simplifies to: \[ x \log y = y - x \]
Rearranging terms to collect all \(x\) parameters on one side: \[ x \log y + x = y \] \[ x(1 + \log y) = y \quad \Rightarrow \quad x = \frac{y}{1 + \log y} \quad \cdots (1) \]


Step 2: Differentiating \(x\) with respect to \(y\) using the Quotient Rule.

Recall the quotient rule for differentiation: \(\frac{d}{dy}\left(\frac{u}{v}\right) = \frac{v \cdot u' - u \cdot v'}{v^2}\). Applying this to equation (1): \[ \frac{dx}{dy} = \frac{(1 + \log y) \cdot \frac{d}{dy}(y) - y \cdot \frac{d}{dy}(1 + \log y)}{(1 + \log y)^2} \] \[ \frac{dx}{dy} = \frac{(1 + \log y) \cdot 1 - y \cdot \left(0 + \frac{1}{y}\right)}{(1 + \log y)^2} \] \[ \frac{dx}{dy} = \frac{1 + \log y - 1}{(1 + \log y)^2} = \frac{\log y}{(1 + \log y)^2} \]


Step 3: Finding \(\frac{dy}{dx}\).

Taking the reciprocal to get the final derivative: \[ \frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} = \frac{(1 + \log y)^2}{\log y} \] Quick Tip: Whenever an implicit function can be explicitly rewritten as \(x = f(y)\), finding \(\frac{dx}{dy}\) first via the quotient rule is often much less prone to algebraic errors than performing implicit differentiation with the product rule directly on \(x \log y = y - x\).

Concept:
For a real-valued function, the domain is the set of all real numbers \(x\) for which the function expression is mathematically well-defined. For square root functions of the form \(\sqrt{g(x)}\), the expression inside the square root must be non-negative: \[ g(x) \geq 0 \]
When multiple square roots are nested, these constraints must be applied systematically to all radical expressions from the inside out, and the final domain is the intersection of all resulting solution sets.


Step 1: Applying the condition for the inner square root.

Consider the inner square root expression \(\sqrt{1 - x^2}\). For this term to be defined: \[ 1 - x^2 \geq 0 \quad \Rightarrow \quad x^2 \leq 1 \]
Taking the square root on both sides defines our first constraint interval: \[ -1 \leq x \leq 1 \quad \Rightarrow \quad x \in [-1, 1] \quad \cdots (1) \]


Step 2: Applying the condition for the outer square root.

For the outer square root to be well-defined, its entire internal expression must be non-negative: \[ x - \sqrt{1 - x^2} \geq 0 \quad \Rightarrow \quad x \geq \sqrt{1 - x^2} \quad \cdots (2) \]
Since a principal square root value is always non-negative (\(\sqrt{1 - x^2} \geq 0\)), equation (2) dynamically forces \(x\) to also be non-negative: \[ x \geq 0 \quad \cdots (3) \]
Combining constraint (1) and condition (3) yields a narrowed temporary boundary: \(x \in [0, 1]\). Now, because both sides of equation (2) are non-negative in this interval, we can safely square both sides without introducing extraneous solution anomalies: \[ x^2 \geq 1 - x^2 \] \[ 2x^2 \geq 1 \quad \Rightarrow \quad x^2 \geq \frac{1}{2} \]
Since we established \(x \geq 0\), taking the positive square root gives: \[ x \geq \frac{1}{\sqrt{2}} \quad \cdots (4) \]


Step 3: Finding the intersection of all constraints.

We find the final domain by taking the intersection of all structural constraint intervals:

From inner radical: \(x \in [-1, 1]\)
From outer radical: \(x \in \left[\frac{1}{\sqrt{2}}, +\infty\right)\)
\[ Domain = [-1, 1] \cap \left[\frac{1}{\sqrt{2}}, +\infty\right) = \left[\frac{1}{\sqrt{2}}, 1\right] \] Quick Tip: Always analyze the sign profiles before squaring an inequality. In this problem, the condition \(x \geq \sqrt{1-x^2}\) immediately eliminates all negative numbers from the domain, allowing you to instantly discard options (1), (2), and (3) without completing any further algebraic calculations!

Concept:
The transpose of a matrix satisfies several key operational properties:

Reversal law for multiplication: \( (M_1 M_2)^T = M_2^T M_1^T \)
Distributive law over addition/subtraction: \( (M_1 \pm M_2)^T = M_1^T \pm M_2^T \)
For a symmetric matrix \(M\), \( M^T = M \)
For a skew-symmetric matrix \(M\), \( M^T = -M \)



Step 1: Determining the nature of matrices X and Y using given conditions.

Since \(A\) and \(B\) are symmetric matrices, we know that: \[ A^T = A \quad and \quad B^T = B \]
Now let's find the transpose of matrix \(X = AB + BA\): \[ X^T = (AB + BA)^T = (AB)^T + (BA)^T \]
Applying the reversal law of transpose: \[ X^T = B^TA^T + A^TB^T \]
Substituting \(A^T = A\) and \(B^T = B\): \[ X^T = BA + AB = AB + BA = X \]
Thus, \(X\) is a **symmetric matrix** (\(X^T = X\)).

Next, let's find the transpose of matrix \(Y = AB - BA\): \[ Y^T = (AB - BA)^T = (AB)^T - (BA)^T \]
Applying the reversal law of transpose: \[ Y^T = B^TA^T - A^TB^T \]
Substituting \(A^T = A\) and \(B^T = B\): \[ Y^T = BA - AB = -(AB - BA) = -Y \]
Thus, \(Y\) is a **skew-symmetric matrix** (\(Y^T = -Y\)).


Step 2: Evaluating the expression \((XY)^T\).

Using the reversal law of transpose on the product expression \((XY)^T\): \[ (XY)^T = Y^T X^T \]
Substitute the results derived in Step 1 (\(X^T = X\) and \(Y^T = -Y\)): \[ (XY)^T = (-Y)(X) = -YX \] Quick Tip: For any two symmetric matrices \(A\) and \(B\) of the same order, the expression \((AB + BA)\) is always symmetric, while the expression \((AB - BA)\) is always skew-symmetric. Remembering this property cuts your problem-solving time in half!

Concept:
A relation \(R\) on a non-empty set \(A\) can be categorized into various properties based on its ordered pairs \((x, y)\):

Reflexive: If \((a, a) \in R\) for every element \(a \in A\).
Symmetric: If \((a, b) \in R \implies (b, a) \in R\) for all \(a, b \in A\).
Transitive: If \((a, b) \in R\) and \((b, c) \in R \implies (a, c) \in R\) for all \(a, b, c \in A\).



Step 1: Writing the relation R in roster form.

The condition for the ordered pairs is \(x + y = 5\), where both elements belong to set \(A = \{1, 2, 3, 4, 5\}\). Let's trace the matching pairs:

If \(x = 1 \implies y = 4\) (since \(1 + 4 = 5\))
If \(x = 2 \implies y = 3\) (since \(2 + 3 = 5\))
If \(x = 3 \implies y = 2\) (since \(3 + 2 = 5\))
If \(x = 4 \implies y = 1\) (since \(4 + 1 = 5\))
If \(x = 5 \implies y = 0\) (not possible because \(0 \notin A\))

Thus, in roster form, the relation is: \[ R = \{(1, 4), (2, 3), (3, 2), (4, 1)\} \]


Step 2: Testing for Reflexive, Symmetric, and Transitive properties.

1. Reflexivity check:
For \(R\) to be reflexive, every element \(a \in A\) must relate to itself, i.e., \((1,1), (2,2), \dots \in R\).
Since \((1, 1) \notin R\) (as \(1 + 1 = 2 \neq 5\)), **\(R\) is not reflexive**.

2. Symmetry check:
We observe the pairs:

\((1, 4) \in R \implies (4, 1) \in R\)
\((2, 3) \in R \implies (3, 2) \in R\)

Since \((x, y) \in R \implies (y, x) \in R\) holds true for all pairs, **\(R\) is symmetric**.

3. Transitivity check:
For \(R\) to be transitive, if \((a, b) \in R\) and \((b, c) \in R\), then \((a, c)\) must also belong to \(R\).
Let's select test pairs from our set: \[ (1, 4) \in R \quad and \quad (4, 1) \in R \]
Here, \(a = 1, b = 4, c = 1\). For transitivity, \((a, c) = (1, 1)\) must belong to \(R\).
However, as seen previously, \((1, 1) \notin R\). Hence, **\(R\) is not transitive**.

Combining these observations, \(R\) is symmetric but neither reflexive nor transitive. Quick Tip: When checking properties on small discrete sets, look for the counter-example first. For transitivity, always check paired cycles like \((a,b)\) and \((b,a)\). If their combined reflexive identity \((a,a)\) is missing from the list, you can instantly conclude that the relation is not transitive.