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JEE 2024 RACE #17 Physics Questions 1 to 7

Physics • Solve question number 1 to 7

Question

1. Find the following integrals: (1) \( \int 3x^7 \, dx \) (2) \( \int 4\sqrt{x} \, dx \) (3) \( \int \frac{dx}{\sqrt{x}} \) (4) \( \int (x^3 - 5x^2 + 7x - 11) \, dx \) (5) \( \int \left( \sqrt[3]{x} - \frac{1}{\sqrt[3]{x}} \right) dx \) (6) \( \int 3x \, dx \) (7) \( \int 5x^2 \, dx \) 2. A particle moves on x-axis as per equation \( x = t^3 - 9t^2 + 15t + 2 \) m. Distance travelled by the particle between \( t=0 \) and \( t=5 \) s is (A) 25 m (B) 39 m (C) 23 m (D) 52 m 3. Suppose velocity of a cricket ball hit by a batsman is given by \( \vec{v} = (2\hat{i} + 6\hat{j}) + (20 - 10t) \hat{k} \) m/s Find the time \( t \) when acceleration of the ball is perpendicular to its velocity. (A) 1 s (B) 2 s (C) 3 s (D) 4 s 4. A particle moves along a straight line and its position as a function of time is given by \( x = t^3 - 3t^2 + 3t + 3 \), then particle (A) stops at \( t=1 \) s and reverses its direction of motion (B) stops at \( t=1 \) s and continues further without change of direction (C) stops at \( t=2 \) s and reverses its direction of motion (D) stops at \( t=2 \) s and continues further without change of direction 5. A ball is thrown vertically upwards with some speed. It reaches two points A and B one after another such that heights of A and B are one fourth and three-fourth of the maximum height attained. If the total time of flight is \( T \), the maximum time taken by the ball to travel from A to B, is: (A) \( \frac{(\sqrt{3}+1)}{4} T \) (B) \( \frac{(\sqrt{3}-1)}{2} T \) (C) \( \frac{(\sqrt{3}+1)}{2} T \) (D) \( \frac{T}{\sqrt{2}} \) 6. The initial velocity of a particle is \( u \) and the acceleration is given by \( kt \), where \( k \) is a positive constant. The distance travelled in time \( t \) is: (A) \( s = ut^2 + kt^2 \) (B) \( s = ut + \frac{kt^3}{6} \) (C) \( s = ut + \frac{kt^2}{2} \) (D) \( s = \frac{ut^2}{2} + \frac{kt^3}{6} \) 7. A motor boat of mass \( m \) moves along a lake with velocity \( V_0 \). At \( t=0 \), the engine of the boat is shut down. Magnitude of resistance force offered to the boat is equal to \( rV \) (\( V \) is instantaneous speed). What is the total distance covered till it stops completely? [Hint: \( F(x) = mV \frac{dV}{dx} = -rV \)] (A) \( \frac{mV_0}{r} \) (B) \( \frac{3mV_0}{2r} \) (C) \( \frac{mV_0}{2r} \) (D) \( \frac{2mV_0}{r} \)

Answer

1. Integrals:

  1. \( \int 3x^7 dx = 3 \times \frac{x^{8}}{8} = \frac{3}{8} x^8 + C \)
  2. \( \int 4\sqrt{x} dx = 4 \int x^{1/2} dx = 4 \times \frac{x^{3/2}}{3/2} = \frac{8}{3} x^{3/2} + C \)
  3. \( \int \frac{dx}{\sqrt{x}} = \int x^{-1/2} dx = 2 x^{1/2} + C \)
  4. \( \int (x^3 - 5x^2 + 7x - 11) dx = \frac{x^4}{4} - \frac{5x^3}{3} + \frac{7x^2}{2} - 11x + C \)
  5. \( \int \left( x^{1/3} - x^{-1/3} \right) dx = \int x^{1/3} dx - \int x^{-1/3} dx = \frac{3}{4} x^{4/3} - \frac{3}{2} x^{2/3} + C \)
  6. \( \int 3x dx = 3 \times \frac{x^2}{2} = \frac{3}{2} x^2 + C \)
  7. \( \int 5x^2 dx = 5 \times \frac{x^3}{3} = \frac{5}{3} x^3 + C \)

2. Distance travelled by particle \( x = t^3 - 9t^2 + 15t + 2 \) from \( t=0 \) to \( t=5 \):

  1. Velocity: \( v = \frac{dx}{dt} = 3t^2 - 18t + 15 \)
  2. Find times when velocity is zero (turning points): \( 3t^2 - 18t + 15 = 0 \Rightarrow t^2 - 6t + 5 = 0 \Rightarrow (t-5)(t-1) = 0 \Rightarrow t=1,5 \)
  3. Positions at \( t=0,1,5 \): \( x(0) = 2 \) \( x(1) = 1 - 9 + 15 + 2 = 9 \) \( x(5) = 125 - 225 + 75 + 2 = -23 \)
  4. Distance travelled = \( |x(1)-x(0)| + |x(5)-x(1)| = |9-2| + |-23-9| = 7 + 32 = 39 \) m

3. Velocity and acceleration of cricket ball:

  1. Velocity: \( \vec{v} = 2\hat{i} + 6\hat{j} + (20 - 10t) \hat{k} \)
  2. Acceleration: \( \vec{a} = \frac{d\vec{v}}{dt} = 0\hat{i} + 0\hat{j} - 10 \hat{k} = -10 \hat{k} \)
  3. Acceleration is perpendicular to velocity when \( \vec{a} \cdot \vec{v} = 0 \): \( \vec{a} \cdot \vec{v} = -10 (20 - 10t) = 0 \Rightarrow 20 - 10t = 0 \Rightarrow t = 2 \) s

4. Particle position \( x = t^3 - 3t^2 + 3t + 3 \):

  1. Velocity: \( v = \frac{dx}{dt} = 3t^2 - 6t + 3 = 3(t^2 - 2t + 1) = 3(t-1)^2 \)
  2. Velocity zero at \( t=1 \) s.
  3. Since velocity is \( 3(t-1)^2 \), it is always \( \geq 0 \) and does not change sign, so particle stops at \( t=1 \) but does not reverse direction.

5. Ball thrown vertically upwards, heights at A and B are \( \frac{1}{4} \) and \( \frac{3}{4} \) of max height, total time of flight \( T \). Find max time from A to B.

Using kinematics, the time to reach height \( h \) is \( t = \frac{T}{2} (1 \pm \sqrt{1 - \frac{h}{H}}) \). For heights \( \frac{1}{4}H \) and \( \frac{3}{4}H \), the time difference between these points is maximum when going up or down. The maximum time between A and B is:

\[ t_{AB} = \frac{(\sqrt{3} + 1)}{4} T \]

6. Particle with initial velocity \( u \) and acceleration \( a = kt \):

  1. Acceleration: \( a = \frac{dv}{dt} = kt \)
  2. Velocity: \( v = u + \int_0^t kt' dt' = u + \frac{kt^2}{2} \)
  3. Distance: \( s = \int_0^t v dt = \int_0^t \left(u + \frac{kt'^2}{2} \right) dt' = ut + \frac{k t^3}{6} \)

7. Motor boat with mass \( m \), initial velocity \( V_0 \), resistance force \( F = -rV \):

  1. Equation: \( m \frac{dV}{dt} = -rV \)
  2. Rewrite as \( m V \frac{dV}{dx} = -rV \Rightarrow m \frac{dV}{dx} = -r \)
  3. Separate variables: \( m dV = -r dx \Rightarrow dx = -\frac{m}{r} dV \)
  4. Total distance till stop (from \( V_0 \) to 0): \( s = \int_0^{s} dx = -\frac{m}{r} \int_{V_0}^0 dV = \frac{m V_0}{r} \)

Final Answer

\boxed{\begin{cases} 1. \int 3x^7 dx = \frac{3}{8} x^8 + C \\ 2. \int 4\sqrt{x} dx = \frac{8}{3} x^{3/2} + C \\ 3. \int \frac{dx}{\sqrt{x}} = 2 x^{1/2} + C \\ 4. \int (x^3 - 5x^2 + 7x - 11) dx = \frac{x^4}{4} - \frac{5x^3}{3} + \frac{7x^2}{2} - 11x + C \\ 5. \int \left( x^{1/3} - x^{-1/3} \right) dx = \frac{3}{4} x^{4/3} - \frac{3}{2} x^{2/3} + C \\ 6. \int 3x dx = \frac{3}{2} x^2 + C \\ 7. \int 5x^2 dx = \frac{5}{3} x^3 + C \\ \text{2. Distance travelled} = 39 \text{ m} \\ \text{3. Time when } \vec{a} \perp \vec{v} = 2 \text{ s} \\ \text{4. Particle stops at } t=1 \text{ s and continues without reversing} \\ \text{5. Max time from A to B} = \frac{(\sqrt{3}+1)}{4} T \\ \text{6. Distance } s = ut + \frac{kt^3}{6} \\ \text{7. Total distance covered} = \frac{m V_0}{r} \end{cases}

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JEE 2024 RACE #17 Physics Questions 1 to 7

1. \( \frac{3}{8} x^8 + C, \frac{8}{3} x^{3/2} + C, 2 x^{1/2} + C, \frac{x^4}{4} - \frac{5x^3}{3} + \frac{7x^2}{2} - 11x + C, \frac{3}{4} x^{4/3} - \frac{3}{2} x^{2/3} + C, \frac{3}{2} x^2 + C, \frac{5}{3} x^3 + C \); 2. 39 m; 3. 2 s; 4. Stops at 1 s, no reversal; 5. \( \frac{(\sqrt{3}+1)}{4} T \); 6. \( s = ut + \frac{kt^3}{6} \); 7. \( \frac{m V_0}{r} \).