PhysicsQ1–2 of 2 questions
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1
PhysicsMediumMCQIIT JEE · 1970
$A$ lift is going up. The variation in the velocity of the lift is given in the graph. What is the height to which the lift takes the passengers? (in $m$)
A
$3.6$
B
$28.8$
C
$36$
D
Cannot be calculated from the above graph
Solution
(C) The total height (displacement) reached by the lift is equal to the area under the velocity-time graph.
The graph is a trapezium with parallel sides of lengths $10 - 2 = 8$ and $12 - 0 = 12$, and a height (velocity) of $3.6$.
Area of trapezium $= \frac{1}{2} \times (\text{sum of parallel sides}) \times (\text{height})$
Area $= \frac{1}{2} \times (8 + 12) \times 3.6$
Area $= \frac{1}{2} \times 20 \times 3.6 = 10 \times 3.6 = 36\,m$.
The graph is a trapezium with parallel sides of lengths $10 - 2 = 8$ and $12 - 0 = 12$, and a height (velocity) of $3.6$.
Area of trapezium $= \frac{1}{2} \times (\text{sum of parallel sides}) \times (\text{height})$
Area $= \frac{1}{2} \times (8 + 12) \times 3.6$
Area $= \frac{1}{2} \times 20 \times 3.6 = 10 \times 3.6 = 36\,m$.
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2
PhysicsEasyMCQIIT JEE · 1970
The ratio between the total intensity of the magnetic field at the equator to the poles is:
A
$1:1$
B
$1:2$
C
$2:1$
D
$1:4$
Solution
(B) The magnetic field of the Earth can be modeled as a magnetic dipole.
At the magnetic equator,the horizontal component of the Earth's magnetic field is $B_e = \frac{\mu_0}{4\pi} \frac{M}{R^3}$.
At the magnetic poles,the vertical component of the Earth's magnetic field is $B_p = \frac{\mu_0}{4\pi} \frac{2M}{R^3}$.
Therefore,the ratio of the intensity at the equator to the poles is $\frac{B_e}{B_p} = \frac{\frac{\mu_0 M}{4\pi R^3}}{\frac{2\mu_0 M}{4\pi R^3}} = \frac{1}{2}$.
Thus,the ratio is $1:2$.
At the magnetic equator,the horizontal component of the Earth's magnetic field is $B_e = \frac{\mu_0}{4\pi} \frac{M}{R^3}$.
At the magnetic poles,the vertical component of the Earth's magnetic field is $B_p = \frac{\mu_0}{4\pi} \frac{2M}{R^3}$.
Therefore,the ratio of the intensity at the equator to the poles is $\frac{B_e}{B_p} = \frac{\frac{\mu_0 M}{4\pi R^3}}{\frac{2\mu_0 M}{4\pi R^3}} = \frac{1}{2}$.
Thus,the ratio is $1:2$.
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