A comprehensive platform for Teachers to create standard question papers and Students to practice Case-Based, Assertion-Reason, and Critical Thinking questions.
Create professional PDF/Word papers with logo, instructions, and mixed question types in minutes.
Explore our repository by Class and Topic. Filter by "Knowledge" or "Competency" levels.
For Students. Take timed MCQ tests to check your understanding. Get instant feedback.
According to NEP 2020, rote learning is out. The focus has shifted to assessing a student's ability to apply concepts in real-life situations.
Questions derived from real-world passages to test analytical skills.
Testing the logic behind concepts, not just the definition.
Open-ended scenarios that require thinking beyond the textbook.
We provide complete AI-Powered Explanations for every question.
Let $V$ be the volume of the cube, $S$ be the surface area, and $x$ be the length of an edge. We are given that $\frac{dV}{dt} = 6~cm^3/s$ and we want to find $\frac{dS}{dt}$ when $x = 8~cm$.
The volume of a cube is given by $V = x^3$, and the surface area is given by $S = 6x^2$.
Differentiating $V = x^3$ with respect to time $t$, we get: $$ \frac{dV}{dt} = 3x^2 \frac{dx}{dt} $$ We are given $\frac{dV}{dt} = 6$, so we have: $$ 6 = 3x^2 \frac{dx}{dt} $$ $$ \frac{dx}{dt} = \frac{6}{3x^2} = \frac{2}{x^2} $$
Differentiating $S = 6x^2$ with respect to time $t$, we get: $$ \frac{dS}{dt} = 12x \frac{dx}{dt} $$
We found that $\frac{dx}{dt} = \frac{2}{x^2}$. Substituting this into the equation for $\frac{dS}{dt}$, we get: $$ \frac{dS}{dt} = 12x \left(\frac{2}{x^2}\right) = \frac{24}{x} $$ Now, we substitute $x = 8~cm$: $$ \frac{dS}{dt} = \frac{24}{8} = 3 $$
The surface area of the cube is increasing at a rate of $3~cm^2/s$.
Final Answer: 3 cm²/s