NEET 2024 Chemistry Question Paper with Answers and Solutions PDF Q3

4ADFK1

• 1.

  A submarine is designed to withstand an absolute pressure of 100 atm. How deep can it go below the water surface? (Consider the density of water = 1000 kg m⁻³, 1 atm = 1 $\times$ 10⁵ Pa and g = 10 m/s²)
  

  • 990 m
    
  • 9000 m
    
  • 99 m
    
  • 9900 m
  View Solution

  ---

• 2.

  A box of mass 15 kg is kept on the floor of a stationary trolley. The coefficient of static friction between the box and the trolley is 0.12. Keeping the box in stationary state over the trolley, the maximum acceleration with which the trolley can be moved horizontally in m s⁻² is: ____.
  

  • 1.2
    
  • 1.8
    
  • 1.5
    
  • 2.1
  View Solution

  ---

• 3.

  The speed of light in vacuum is taken as unity. If light takes 6 min 40 s to reach the Earth from the Sun, the distance between the Sun and the Earth in new unit is: ____.
  

  • 500
    
  • 3 $\times$ 10⁸
    
  • 400
    
  • 3 $\times$ 10¹⁰
  View Solution

  ---

• 4.

  The figure given below shows a long straight solid wire of circular cross-section of radius ‘a’ carrying steady current I. The current I is uniformly distributed across its cross-section. The plot which correctly represents the variation of magnetic field (B) with distance (r) from the axis of the conductor in the region is: ____.
  

  

  • A
  • B
  • C
  • D
  View Solution

  ---

• 5.

  When \(1\ dm^3\) of \(CO_2\) gas is passed over hot coke, the volume of gaseous mixture after complete reaction at STP becomes \(1.4\ dm^3\). The composition of the gaseous mixture at STP is:

  • \(0.6\ dm^3\) of \(CO\), \(0.8\ dm^3\) of \(CO_2\)
    
  • \(0.6\ dm^3\) of \(CO\), \(0.9\ dm^3\) of \(CO_2\)
    
  • \(0.8\ dm^3\) of \(CO\), \(0.6\ dm^3\) of \(CO_2\)
    
  • \(0.8\ dm^3\) of \(CO\), \(0.7\ dm^3\) of \(CO_2\)
  View Solution

  ---

• 6.

  The number of hydrogen atoms present in \(5.4\text{ g}\) of urea is: \[ \text{Given: Molar mass of urea }=60\text{ g mol}^{-1} \] \[ N_A=6.022\times10^{23}\text{ particles mol}^{-1} \]

  • \(2.168\times10^{22}\)
    
  • \(2.168\times10^{23}\)
    
  • \(1.084\times10^{22}\)
    
  • \(1.084\times10^{23}\)
  View Solution