AP EAPCET 2026 May 13 Shift 1 Engineering Question Paper Available- Download Solution PDF

Concept:


According to Lagrange's Mean Value Theorem, if a function \(f(x)\) is continuous on \([a,b]\) and differentiable on \((a,b)\), then there exists at least one point \(c\in(a,b)\) such that:

\(\displaystyle f'(c)=\frac{f(b)-f(a)}{b-a}\)



Step 1: Write the given function and interval.


Here,

\(\displaystyle f(x)=\sqrt{25-x^2}\)

and the interval is:

\(\displaystyle [1,5]\)

So,

\(\displaystyle a=1,\quad b=5\)



Step 2: Find \(f(1)\) and \(f(5)\).


\(\displaystyle f(1)=\sqrt{25-1^2}\)

\(\displaystyle f(1)=\sqrt{24}=2\sqrt{6}\)

Also,

\(\displaystyle f(5)=\sqrt{25-5^2}\)

\(\displaystyle f(5)=\sqrt{0}=0\)



Step 3: Find the average rate of change.


\(\displaystyle \frac{f(5)-f(1)}{5-1}=\frac{0-2\sqrt{6}}{4}\)

\(\displaystyle =-\frac{\sqrt{6}}{2}\)



Step 4: Differentiate the function.


\(\displaystyle f(x)=\sqrt{25-x^2}=(25-x^2)^{1/2}\)

\(\displaystyle f'(x)=\frac{1}{2}(25-x^2)^{-1/2}(-2x)\)

\(\displaystyle f'(x)=\frac{-x}{\sqrt{25-x^2}}\)



Step 5: Apply Lagrange's Mean Value Theorem.


\(\displaystyle f'(c)=-\frac{\sqrt{6}}{2}\)

So,

\(\displaystyle \frac{-c}{\sqrt{25-c^2}}=-\frac{\sqrt{6}}{2}\)

Removing the negative sign,

\(\displaystyle \frac{c}{\sqrt{25-c^2}}=\frac{\sqrt{6}}{2}\)



Step 6: Square both sides.


\(\displaystyle \frac{c^2}{25-c^2}=\frac{6}{4}\)

\(\displaystyle \frac{c^2}{25-c^2}=\frac{3}{2}\)

Cross multiply:

\(\displaystyle 2c^2=3(25-c^2)\)

\(\displaystyle 2c^2=75-3c^2\)

\(\displaystyle 5c^2=75\)

\(\displaystyle c^2=15\)

\(\displaystyle c=\sqrt{15}\)



Hence,

\(\displaystyle \boxed{c=\sqrt{15}}\) Quick Tip: For Lagrange's Mean Value Theorem, always use: \[ f'(c)=\frac{f(b)-f(a)}{b-a} \] Then solve for \(c\) inside the open interval \((a,b)\).

Concept:


For integrals containing powers of \((x-1)\) and \((x+2)\), a useful substitution is:

\(\displaystyle u=\frac{x-1}{x+2}\)

This is useful because the derivative of this fraction produces a factor involving \((x+2)^2\).



Step 1: Rewrite the integral.


\(\displaystyle I=\int \frac{1}{\sqrt[4]{(x-1)^3(x+2)^5}}\,dx\)

This can be written as:

\(\displaystyle I=\int (x-1)^{-\frac{3}{4}}(x+2)^{-\frac{5}{4}}\,dx\)



Step 2: Choose a suitable expression.


Let us check the derivative of:

\(\displaystyle \left(\frac{x-1}{x+2}\right)^{\frac{1}{4}}\)



Step 3: Differentiate the inner fraction.


Let,

\(\displaystyle u=\frac{x-1}{x+2}\)

Then,

\(\displaystyle \frac{du}{dx}=\frac{(x+2)(1)-(x-1)(1)}{(x+2)^2}\)

\(\displaystyle \frac{du}{dx}=\frac{x+2-x+1}{(x+2)^2}\)

\(\displaystyle \frac{du}{dx}=\frac{3}{(x+2)^2}\)



Step 4: Differentiate the complete expression.


\(\displaystyle \frac{d}{dx}\left[\left(\frac{x-1}{x+2}\right)^{\frac{1}{4}}\right]\)

\(\displaystyle =\frac{1}{4}\left(\frac{x-1}{x+2}\right)^{-\frac{3}{4}}\cdot \frac{3}{(x+2)^2}\)

\(\displaystyle =\frac{3}{4}(x-1)^{-\frac{3}{4}}(x+2)^{\frac{3}{4}}\cdot (x+2)^{-2}\)

\(\displaystyle =\frac{3}{4}(x-1)^{-\frac{3}{4}}(x+2)^{-\frac{5}{4}}\)



Step 5: Compare with the given integrand.


The given integrand is:

\(\displaystyle (x-1)^{-\frac{3}{4}}(x+2)^{-\frac{5}{4}}\)

We obtained:

\(\displaystyle \frac{d}{dx}\left[\left(\frac{x-1}{x+2}\right)^{\frac{1}{4}}\right]=\frac{3}{4}(x-1)^{-\frac{3}{4}}(x+2)^{-\frac{5}{4}}\)

Therefore,

\(\displaystyle (x-1)^{-\frac{3}{4}}(x+2)^{-\frac{5}{4}}=\frac{4}{3}\frac{d}{dx}\left[\left(\frac{x-1}{x+2}\right)^{\frac{1}{4}}\right]\)



Step 6: Write the final integral.


\(\displaystyle I=\frac{4}{3}\left(\frac{x-1}{x+2}\right)^{\frac{1}{4}}+C\)

Hence,

\(\displaystyle \boxed{\frac{4}{3}\left(\frac{x-1}{x+2}\right)^{\frac{1}{4}}+C}\) Quick Tip: When an integral contains powers of \((x-1)\) and \((x+2)\), try differentiating \(\left(\frac{x-1}{x+2}\right)^n\).

Concept:


Surface tension is defined as force acting per unit length.

So,

\(\displaystyle Surface tension=\frac{Force}{Length}\)



Step 1: Find the dimensional formula of surface tension.


Force has dimensional formula:

\(\displaystyle [F]=[MLT^{-2}]\)

Length has dimensional formula:

\(\displaystyle [L]\)

Therefore,

\(\displaystyle [Surface tension]=\frac{[MLT^{-2}]}{[L]}\)

\(\displaystyle =[MT^{-2}]\)



Step 2: Check Energy per Area.


Energy has dimensional formula:

\(\displaystyle [E]=[ML^2T^{-2}]\)

Area has dimensional formula:

\(\displaystyle [A]=[L^2]\)

So,

\(\displaystyle \left[\frac{Energy}{Area}\right]=\frac{[ML^2T^{-2}]}{[L^2]}\)

\(\displaystyle =[MT^{-2}]\)



Step 3: Compare both dimensions.


Surface tension:

\(\displaystyle [MT^{-2}]\)

Energy per area:

\(\displaystyle [MT^{-2}]\)

Both are the same.



Step 4: Final conclusion.


Hence, the correct answer is:

\(\displaystyle \boxed{Energy / Area}\) Quick Tip: Surface tension can be written as force per length or energy per area. Both give the same dimensional formula \([MT^{-2}]\).

Concept:


To convert speed from \(km/h\) to \(m/s\), multiply by:

\(\displaystyle \frac{5}{18}\)



Step 1: Write the given speed.


\(\displaystyle 54\ km/h\)



Step 2: Convert into \(m/s\).


\(\displaystyle 54\ km/h=54\times \frac{5}{18}\ m/s\)

\(\displaystyle =3\times 5\ m/s\)

\(\displaystyle =15\ m/s\)



Step 3: Final conclusion.


Hence,

\(\displaystyle \boxed{54\ km/h=15\ m/s}\) Quick Tip: To convert \(km/h\) into \(m/s\), multiply by \(\frac{5}{18}\). To convert \(m/s\) into \(km/h\), multiply by \(\frac{18}{5}\).

Concept:


In addition and subtraction, the final answer should be rounded according to the least number of decimal places among the given numbers.



Step 1: Add the given numbers.


\(\displaystyle 3.456+2.1=5.556\)



Step 2: Check decimal places.


The number \(3.456\) has three decimal places.

The number \(2.1\) has one decimal place.

So, the final answer must be rounded to one decimal place.



Step 3: Round the answer.


\(\displaystyle 5.556\)

Rounded to one decimal place:

\(\displaystyle 5.6\)



Step 4: Final conclusion.


Hence, the correct result is:

\(\displaystyle \boxed{5.6}\) Quick Tip: For addition and subtraction, round the final answer according to the least number of decimal places, not the least number of significant figures.

Concept:


Percentage error is calculated using the formula:

\(\displaystyle Percentage Error=\frac{Absolute Error}{True Value}\times 100\)



Step 1: Write the given values.


Measured value:

\(\displaystyle 98\)

True value:

\(\displaystyle 100\)



Step 2: Find absolute error.


\(\displaystyle Absolute Error=|True Value-Measured Value|\)

\(\displaystyle =|100-98|\)

\(\displaystyle =2\)



Step 3: Find percentage error.


\(\displaystyle Percentage Error=\frac{2}{100}\times 100\)

\(\displaystyle =2%\)



Step 4: Final conclusion.


Hence, the percentage error is:

\(\displaystyle \boxed{2%}\) Quick Tip: Percentage error is always calculated by dividing absolute error by true value and then multiplying by \(100\).

Concept:


Least count of a screw gauge is given by:

\(\displaystyle Least Count=\frac{Pitch}{Number of divisions on circular scale}\)



Step 1: Write the given values.


Pitch:

\(\displaystyle 0.5\ mm\)

Number of circular scale divisions:

\(\displaystyle 50\)



Step 2: Apply the formula.


\(\displaystyle Least Count=\frac{0.5}{50}\ mm\)

\(\displaystyle =0.01\ mm\)



Step 3: Final conclusion.


Hence, the least count is:

\(\displaystyle \boxed{0.01\ mm}\) Quick Tip: For screw gauge: \[ Least Count=\frac{Pitch}{Number of circular scale divisions} \]

Concept:


Displacement from a velocity-time graph is equal to the area under the velocity-time graph.

Here, the velocity-time graph is a straight line from \(0\) to \(20\ m/s\), so the area is triangular.



Step 1: Identify base and height of the triangle.


Base of the triangle:

\(\displaystyle 10\ s\)

Height of the triangle:

\(\displaystyle 20\ m/s\)



Step 2: Use area of triangle.


\(\displaystyle Displacement=\frac{1}{2}\times base\times height\)

\(\displaystyle =\frac{1}{2}\times 10\times 20\)

\(\displaystyle =100\ m\)



Step 3: Final conclusion.


Hence, the displacement in \(10\ s\) is:

\(\displaystyle \boxed{100\ m}\) Quick Tip: In a velocity-time graph, displacement is always equal to the area under the graph.

Concept:


Acceleration is the rate of change of velocity with respect to time.

So,

\(\displaystyle a=\frac{dv}{dt}\)



Step 1: Write the given velocity equation.


\(\displaystyle v=4t\)



Step 2: Differentiate velocity with respect to time.


\(\displaystyle a=\frac{dv}{dt}\)

\(\displaystyle a=\frac{d}{dt}(4t)\)

\(\displaystyle a=4\)



Step 3: Write the unit.


Acceleration is measured in:

\(\displaystyle m/s^2\)

So,

\(\displaystyle a=4\ m/s^2\)



Step 4: Final conclusion.


Hence, the acceleration is:

\(\displaystyle \boxed{4\ m/s^2}\) Quick Tip: If velocity is given as a function of time, acceleration is found by differentiating velocity with respect to time.

Concept:


When two bodies move in the same direction, their relative velocity is the difference of their velocities.



Step 1: Write the velocities.


Velocity of first car:

\(\displaystyle v_1=20\ m/s\)

Velocity of second car:

\(\displaystyle v_2=30\ m/s\)



Step 2: Find relative velocity.


Since both cars are moving in the same direction:

\(\displaystyle v_{relative}=|v_2-v_1|\)

\(\displaystyle =|30-20|\)

\(\displaystyle =10\ m/s\)



Step 3: Final conclusion.


Hence, the velocity of one car relative to the other is:

\(\displaystyle \boxed{10\ m/s}\) Quick Tip: For same direction, relative velocity is the difference of velocities. For opposite direction, relative velocity is the sum of velocities.

Concept:


For uniformly accelerated motion, distance covered is given by:

\(\displaystyle s=ut+\frac{1}{2}at^2\)

where,

\(u=\) initial velocity

\(a=\) acceleration

\(t=\) time



Step 1: Write the given values.


The body starts from rest, so:

\(\displaystyle u=0\)

Acceleration:

\(\displaystyle a=2\ m/s^2\)

Time:

\(\displaystyle t=5\ s\)



Step 2: Apply the formula.


\(\displaystyle s=ut+\frac{1}{2}at^2\)

\(\displaystyle s=(0)(5)+\frac{1}{2}(2)(5)^2\)

\(\displaystyle s=0+1\times 25\)

\(\displaystyle s=25\ m\)



Step 3: Final conclusion.


Hence, the distance covered is:

\(\displaystyle \boxed{25\ m}\) Quick Tip: If a body starts from rest, then \(u=0\), and the formula becomes \(s=\frac{1}{2}at^2\).

Concept:


For uniformly accelerated or retarded motion, we use:

\(\displaystyle v^2=u^2+2as\)

where,

\(u=\) initial velocity

\(v=\) final velocity

\(a=\) acceleration

\(s=\) distance



Step 1: Write the given values.


Initial velocity:

\(\displaystyle u=20\ m/s\)

Final velocity:

\(\displaystyle v=0\ m/s\)

Retardation:

\(\displaystyle a=-5\ m/s^2\)



Step 2: Apply the formula.


\(\displaystyle v^2=u^2+2as\)

Substitute the values:

\(\displaystyle 0^2=20^2+2(-5)s\)

\(\displaystyle 0=400-10s\)

\(\displaystyle 10s=400\)

\(\displaystyle s=40\ m\)



Step 3: Final conclusion.


Hence, the stopping distance is:

\(\displaystyle \boxed{40\ m}\) Quick Tip: For stopping distance, final velocity is \(0\), and retardation is taken as negative acceleration.

Concept:


For a projectile projected with speed \(u\) at an angle \(\theta\), the horizontal range is given by:

\(\displaystyle R=\frac{u^2\sin 2\theta}{g}\)

where,

\(R=\) range of projectile

\(u=\) initial speed

\(\theta=\) angle of projection

\(g=\) acceleration due to gravity



Step 1: Write the given values.


Initial speed:

\(\displaystyle u=20\ m/s\)

Angle of projection:

\(\displaystyle \theta=30^\circ\)

Acceleration due to gravity:

\(\displaystyle g=10\ m/s^2\)



Step 2: Use the range formula.


\(\displaystyle R=\frac{u^2\sin 2\theta}{g}\)

Substitute the values:

\(\displaystyle R=\frac{(20)^2\sin(2\times 30^\circ)}{10}\)

\(\displaystyle R=\frac{400\sin 60^\circ}{10}\)



Step 3: Use the value of \(\sin 60^\circ\).


\(\displaystyle \sin 60^\circ=\frac{\sqrt{3}}{2}\)

So,

\(\displaystyle R=\frac{400}{10}\times \frac{\sqrt{3}}{2}\)

\(\displaystyle R=40\times \frac{\sqrt{3}}{2}\)

\(\displaystyle R=20\sqrt{3}\)



Step 4: Find approximate value.


Since,

\(\displaystyle \sqrt{3}\approx 1.732\)

Therefore,

\(\displaystyle R=20\times 1.732\)

\(\displaystyle R=34.64\ m\)



Step 5: Choose the nearest option.


The nearest option to \(34.64\ m\) is:

\(\displaystyle 30\ m\)

Hence, the correct answer is:

\(\displaystyle \boxed{30\ m}\) Quick Tip: For projectile range, use: \[ R=\frac{u^2\sin 2\theta}{g} \] Always remember that the angle becomes \(2\theta\) inside the sine function.

Concept:


When two vectors are perpendicular to each other, the angle between them is:

\(\displaystyle 90^\circ\)

For two perpendicular vectors, the magnitude of resultant is found using Pythagoras theorem:

\(\displaystyle R=\sqrt{A^2+B^2}\)

where,

\(A\) and \(B\) are the magnitudes of the two vectors.



Step 1: Write the given vector magnitudes.


First vector magnitude:

\(\displaystyle A=3\)

Second vector magnitude:

\(\displaystyle B=4\)



Step 2: Use resultant formula for perpendicular vectors.


Since the vectors are perpendicular,

\(\displaystyle R=\sqrt{A^2+B^2}\)

Substitute the values:

\(\displaystyle R=\sqrt{3^2+4^2}\)



Step 3: Simplify the expression.


\(\displaystyle R=\sqrt{9+16}\)

\(\displaystyle R=\sqrt{25}\)

\(\displaystyle R=5\)



Step 4: Final conclusion.


Hence, the resultant of the two perpendicular vectors is:

\(\displaystyle \boxed{5}\) Quick Tip: For perpendicular vectors, use Pythagoras theorem: \[ R=\sqrt{A^2+B^2} \] The pair \(3,4,5\) is a common right-triangle result.

Concept:


For crossing a river in minimum time, the boat should be directed perpendicular to the river flow.

In minimum-time crossing, only the velocity component of the boat perpendicular to the river bank is used for crossing the width.

So,

\(\displaystyle Minimum time=\frac{Width of river}{Speed of boat in still water}\)



Step 1: Write the given values.


Speed of boat in still water:

\(\displaystyle v_b=5\ m/s\)

Speed of river:

\(\displaystyle v_r=3\ m/s\)

Width of river:

\(\displaystyle d=40\ m\)



Step 2: Understand the role of river speed.


For minimum time, the boat is aimed straight across the river.

The river current only causes drift downstream.

It does not affect the time needed to cross the width.

Therefore, for minimum time, we use only:

\(\displaystyle v_b=5\ m/s\)



Step 3: Apply the formula.


\(\displaystyle t_{\min}=\frac{d}{v_b}\)

\(\displaystyle t_{\min}=\frac{40}{5}\)

\(\displaystyle t_{\min}=8\ s\)



Step 4: Final conclusion.


Hence, the minimum time required is:

\(\displaystyle \boxed{8\ s}\) Quick Tip: For minimum time to cross a river, point the boat perpendicular to the bank. The river current affects drift, not the crossing time.