JEE MAIN - Physics (2021 - 16th March Evening Shift - No. 4)
Statement I : A cyclist is moving on an unbanked road with a speed of 7 kmh$$-$$1 and takes a sharp circular turn along a path of radius of 2m without reducing the speed. The static friction coefficient is 0.2. The cyclist will not slip and pass the curve. (g = 9.8 m/s2)
Statement II : If the road is banked at an angle of 45$$^\circ$$, cyclist can cross the curve of 2m radius with the speed of 18.5 kmh$$-$$1 without slipping.
In the light of the above statements, choose the correct answer from the options given below.
Statement II : If the road is banked at an angle of 45$$^\circ$$, cyclist can cross the curve of 2m radius with the speed of 18.5 kmh$$-$$1 without slipping.
In the light of the above statements, choose the correct answer from the options given below.
Statement I is incorrect and statement II is correct
Both statement I and statement II are true
Statement I is correct and statement II is incorrect
Both statement I and statement II are false
Explanation
On a horizontal ground,
$${v_{\max }} = \sqrt {\mu Rg} = \sqrt {0.2 \times 2 \times 9.8} = 1.97$$ m/s
= $$1.97 \times {{18} \over 5}$$
= 7.12 km/hr = 7.2 km/hr
Statement - 2
$${v_{\max }} = \sqrt {gr\left( {{{\tan \theta + \mu } \over {1 - \mu \tan \theta }}} \right)} = \sqrt {2 \times 9.8 \times {{12} \over {0.8}}} = 19.5$$ km/hr
$${v_{\min }} = \sqrt {rg\left( {{{\tan \theta - \mu } \over {1 + \mu \tan \theta }}} \right)} = \sqrt {2 \times 9.8 \times {{0.8} \over {1.2}}} = 12.01$$ km/hr
$${v_{\max }} = \sqrt {\mu Rg} = \sqrt {0.2 \times 2 \times 9.8} = 1.97$$ m/s
= $$1.97 \times {{18} \over 5}$$
= 7.12 km/hr = 7.2 km/hr
Statement - 2
$${v_{\max }} = \sqrt {gr\left( {{{\tan \theta + \mu } \over {1 - \mu \tan \theta }}} \right)} = \sqrt {2 \times 9.8 \times {{12} \over {0.8}}} = 19.5$$ km/hr
$${v_{\min }} = \sqrt {rg\left( {{{\tan \theta - \mu } \over {1 + \mu \tan \theta }}} \right)} = \sqrt {2 \times 9.8 \times {{0.8} \over {1.2}}} = 12.01$$ km/hr
Comments (0)
