The position of both an electron and helium atom is known within . The momentum of the electron is known within . The minimum uncertainty in the measurement of the momentum of the helium atom is
1
2
3
4
Official Solution
Correct Option: (2)
The Heisenberg uncertainty principle,
, where Uncertainty in position,
Uncertainty in momentum and constant.
As is same for electron and helium and is a constant, therefore minimum uncertainty in the measurement of the momentum of the helium atom will be same as tlat of an electron which is
02
PYQ 2018
medium
physicsID: neet-ug-
A moving block having mass m, collides with another stationary block having mass 4m. The lighter block comes to rest after collision. When the initial velocity of the lighter block is v, then the value of coefficient of restitution (e) will be
1
0.5
2
0.8
3
0.25
4
0.4
Official Solution
Correct Option: (3)
The correct option is (C): 0.25
03
PYQ 2022
medium
physicsID: neet-ug-
Two objects of mass 10 kg and 20 kg respectively are connected to the two ends of a rigid rod of length 10 m with negligible mass. The distance of the center of mass of the system from the 10 kg mass is:
1
2
3
4
Official Solution
Correct Option: (2)
To find the center of mass of the system consisting of two objects attached to a rigid rod, we use the formula for the center of mass of a system of particles:
where is the mass and is the position of each particle. Given: Mass of first object
Mass of second object
Length of the rod
Placing the 10 kg mass at the origin (0 m) and the 20 kg mass at 10 m, the positions are:
for 10 kg mass,
for 20 kg mass.
Thus, the center of mass is calculated as:
Thus, the distance from the 10 kg mass to the center of mass is .
04
PYQ 2022
hard
physicsID: neet-ug-
A shell of mass m is at rest initially. It explodes into three fragments having mass in the ratio . If the fragments having equal mass fly off along mutually perpendicular directions with speed v, the speed of the third (lighter) fragment is
1
2
3
4
Official Solution
Correct Option: (3)
To solve this problem, let's use the law of conservation of momentum. Initially, the shell is at rest, so its total initial momentum is zero. After the explosion, the shell breaks into three fragments with their mass ratios . Let's denote the mass of the third fragment as and the common mass of the other two heavier fragments as .
The fragments and move along perpendicular directions with speed . We need to calculate the speed of the third fragment .
To use the conservation of momentum, the total momentum in each direction should remain zero.
Vector sum of momenta in the x and y-directions:
Let's assume moves along the x-axis and moves along the y-axis:
Along the x-axis:
Along the y-axis:
Since , substitute these into the momentum equations:
The speed of the third fragment is given by combining the components:
