Exam Questions & Step-by-Step Markschemes (6 Problems)
Question 1 • 2024
4 Marks
An open box is to be made from a square sheet of tin of side 18\text{ cm} by cutting off equal squares from each corner and turning up the sides. What should be the side of the square to be cut off so that the volume of the box is maximum?
Answer: x = 3\text{ cm} \quad (V_{\max} = 432\text{ cm}^3)
Question 2 • 2024
4 Marks
Show that the right circular cylinder of maximum volume that can be inscribed in a sphere of fixed radius R has height h = \frac{2R}{\sqrt{3}} , and find this maximum volume.
Answer: h = \frac{2R}{\sqrt{3}}, \quad V_{\max} = \frac{4\pi R^3}{3\sqrt{3}}
Question 3 • 2024
4 Marks
Show that the volume of the greatest cylinder that can be inscribed in a sphere of radius R is \frac{4\pi R^3}{3\sqrt{3}} .
Answer: V_{\max} = \frac{4\pi R^3}{3\sqrt{3}}
Question 4 • 2024
4 Marks
A ladder 5\text{ m} long is leaning against a vertical wall. The bottom of the ladder is pulled along the ground away from the wall at the rate of 2\text{ m/s} . How fast is its height on the wall decreasing when the foot of the ladder is 4\text{ m} away from the wall?
Answer: \text{Height is decreasing at } \frac{8}{3}\text{ m/s} \approx 2.67\text{ m/s}
Question 5 • 2024
3 Marks
એક ગોળાકાર ફુગ્ગામાં હવા ભરતાં તેની ત્રિજ્યા 3\text{ cm/s} ના દરે વધે છે. જ્યારે ત્રિજ્યા 10\text{ cm} હોય ત્યારે તેના ઘનફળના વધવાનો દર શોધો. (The radius of a spherical balloon increases at 3\text{ cm/s} . Find the rate of increase of its volume when the radius is 10\text{ cm} .)
Answer: \frac{dV}{dt} = 1200\pi \text{ cm}^3/\text{s}
Question 6 • 2023
4 Marks
Water is dripping out from a conical funnel at a uniform rate of 4\text{ cm}^3/\text{sec} through a tiny hole at the vertex in the bottom. When the slant height of the water is 3\text{ cm} , find the rate of decrease of the slant height of the water cone, given that the semi-vertical angle of the funnel is 30^\circ .
Answer: \frac{dl}{dt} = -\frac{32\sqrt{3}}{27\pi} \text{ cm/sec}