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The questions your board exam loves to ask

800 most-asked Class 11 & 12 (+1 / +2) questions across Physics, Chemistry, Maths and Biology — each with a model answer and the exact marking-scheme points examiners reward. Revise smart, walk in calm.

800 board questionsModel answersMarking-scheme pointsEvery chapterCBSE · ISC · State boards

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PhysicsClass 115 marksmedium

Motion in a Plane

A projectile is launched with speed u at an angle theta to the horizontal. Derive expressions for its time of flight, maximum height and horizontal range.

Reveal model answer + marking points

Resolve u into ux = u cos(theta) and uy = u sin(theta). Vertical motion has acceleration -g. Time of flight: uy - g(T/2) = 0 at the top, so T = 2u sin(theta)/g. Maximum height: H = uy^2/2g = u^2 sin^2(theta)/2g. Horizontal range: R = ux * T = u cos(theta) * 2u sin(theta)/g = u^2 sin(2 theta)/g. R is maximum when theta = 45 degrees.

T = 2u sin(theta)/g ; H = u^2 sin^2(theta)/2g ; R = u^2 sin(2theta)/g

Marking-scheme points

  • Resolve into horizontal and vertical components
  • T = 2u sin(theta)/g
  • H = u^2 sin^2(theta)/2g
  • R = u^2 sin(2 theta)/g
  • R max at theta = 45 deg
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PhysicsClass 115 markshard

Thermodynamics

Describe the four steps of a Carnot cycle and write the expression for the efficiency of a Carnot engine.

Reveal model answer + marking points

A Carnot cycle has four reversible steps: (1) isothermal expansion at the source temperature T_h (heat Q_h absorbed); (2) adiabatic expansion (temperature falls from T_h to T_c); (3) isothermal compression at the sink temperature T_c (heat Q_c rejected); (4) adiabatic compression (temperature rises from T_c back to T_h). The efficiency is eta = 1 - Q_c/Q_h = 1 - T_c/T_h, where temperatures are in kelvin. Efficiency depends only on the two temperatures and is always less than 1.

eta = 1 - T_c/T_h

Marking-scheme points

  • 4 steps: isothermal exp, adiabatic exp, isothermal comp, adiabatic comp
  • eta = 1 - T_c/T_h
  • T in kelvin; eta < 1 always
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MathsClass 115 markshard

Trigonometric Functions

Prove that cos A/(1 - tan A) + sin A/(1 - cot A) = sin A + cos A.

Reveal model answer + marking points

Write tan A = sin A/cos A and cot A = cos A/sin A. First term: cos A/(1 - sin A/cos A) = cos A/((cos A - sin A)/cos A) = cos^2 A/(cos A - sin A). Second term: sin A/(1 - cos A/sin A) = sin A/((sin A - cos A)/sin A) = sin^2 A/(sin A - cos A) = -sin^2 A/(cos A - sin A). Adding: [cos^2 A - sin^2 A]/(cos A - sin A) = [(cos A - sin A)(cos A + sin A)]/(cos A - sin A) = cos A + sin A. Hence proved.

cos^2 A - sin^2 A = (cosA - sinA)(cosA + sinA)

Marking-scheme points

  • Convert tan A and cot A into sin/cos
  • First term = cos^2 A/(cosA - sinA); second = -sin^2 A/(cosA - sinA)
  • Add: (cos^2 A - sin^2 A)/(cosA - sinA) = cosA + sinA
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MathsClass 115 markshard

Sequences and Series

Find the sum of the series 1 x 2 + 2 x 3 + 3 x 4 + ... up to n terms.

Reveal model answer + marking points

The rth term is a_r = r(r + 1) = r^2 + r. So the sum S = sum over r of (r^2 + r) = sum of r^2 + sum of r. Using sum of r^2 = n(n + 1)(2n + 1)/6 and sum of r = n(n + 1)/2: S = n(n + 1)(2n + 1)/6 + n(n + 1)/2. Take n(n + 1)/6 common: S = [n(n + 1)/6][(2n + 1) + 3] = [n(n + 1)/6](2n + 4) = [n(n + 1) x 2(n + 2)]/6 = n(n + 1)(n + 2)/3.

S = n(n + 1)(n + 2)/3

Marking-scheme points

  • rth term = r(r + 1) = r^2 + r
  • Sum = sum of r^2 + sum of r
  • = n(n + 1)(2n + 1)/6 + n(n + 1)/2 = n(n + 1)(n + 2)/3
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MathsClass 115 markshard

Straight Lines

Find the image (reflection) of the point (3, 8) in the line x + 3y - 7 = 0.

Reveal model answer + marking points

For the image (h, k) of point (x1, y1) in line Ax + By + C = 0, we use (h - x1)/A = (k - y1)/B = -2(A x1 + B y1 + C)/(A^2 + B^2). Here A = 1, B = 3, C = -7, (x1, y1) = (3, 8). A x1 + B y1 + C = 3 + 24 - 7 = 20, and A^2 + B^2 = 1 + 9 = 10, so the common ratio = -2(20)/10 = -4. Then h = 3 + 1(-4) = -1 and k = 8 + 3(-4) = 8 - 12 = -4. Hence the image is (-1, -4).

(h - x1)/A = (k - y1)/B = -2(A x1 + B y1 + C)/(A^2 + B^2)

Marking-scheme points

  • (h - x1)/A = (k - y1)/B = -2(A x1 + B y1 + C)/(A^2 + B^2)
  • Ratio = -2(20)/10 = -4
  • Image = (-1, -4)
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