एक गेंद को एक चिकने अर्ध-गोलाकार बर्तन के बिंदु P से आराम से छोड़ा जाता है जैसा कि चित्र में दिखाया गया है। बिंदु Q पर गेंद पर केन्द्रापसारक बल और सामान्य प्रतिक्रिया का अनुपात A है जबकि बिंदु P के संबंध में बिंदु Q की कोणीय स्थिति α है। निम्नलिखित में से कौन सा ग्राफ A और α के बीच सही संबंध का प्रतिनिधित्व करता है जब गेंद क्यू से आर?

Answer (Detailed Solution Below)

Option 3 :
Free
JEE Main 04 April 2024 Shift 1
90 Qs. 300 Marks 180 Mins

Detailed Solution

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CONCEPT:

  • Centripetal acceleration is defined as the ratio of the square of the velocity and radius and it is written as,

           ac =  

           here we have ac as the centripetal acceleration, v is the velocity and R is the radius.

CALCULATION:

A ball is released from rest from point P of a smooth semi-spherical vessel as shown in the figure below,


Using the trigonometry identity to find h we have;

⇒ h = R sin

and the first law of motion we have;

v2 - u2 = 2gh

Here is the initial velocity, u = 0 m/s we have;

v2 = 2gh

⇒ v =  

⇒ v =  -----(1)

The centripetal acceleration is written as;

ac =  

Now, on putting the value of equation (1) above we have,

ac =  

⇒ ac = 2g sin -----(2)

At point Q we can see that normal N and force F which mg sin we have;

N -  mg sin α  = mac

⇒ N = mac + mg sinα ----(3)

Now, on putting the value of equation (2) in equation (3) we have;

N = m × 2 g sin α + mg sin α 

N = 3 × mg sin α 

The ratio of the centripetal force and normal reaction is written as;

⇒  = constant

Therefore, the ratio of the centripetal force and normal reaction is constant.

Hence, option (3) is the correct answer.

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