Contest Problems: ΔT, mvt2/r, Tb-mg, Solar Eclipse Times & V-emf

[sparkle123]

The following questions aren't for homework, just contest preparation. Thanks for your help!

1. A ball of mass m attached to an inextensible string of length R in swung around a vertical circle just fast enough so that the string is always fully stretched. Let ΔT denote the difference between the tension in the string at the bottom and at the top of the circle, v_b and v_t the speed of the ball at the bottom and at the top, respectively. Then, taking “dependence” to be with respect to a set of independent variables,
Ans: ΔT is independent of R, v_b and v_t.

My Work:
mv_t²/r = T_t + mg
mv_b²/r = T_b - mg


2. An object of mass m hangs motionless from a vertical spring. When the object is pulled down to a new rest position, the total mechanical energy of the system
Ans: increases


3. To a good approximation, Earth and Jupiter move around the
Sun in circular orbits of 1.49 x 108 km and 7.79 x 109 km
radius, respectively. What is the maximum error that can
arise in the prediction of solar eclipse times (as observed from
Earth) on Jupiter caused by one of its moons if one fails to
take into account the variation of the relative position of the
two planets?
Answer: 2.6 x 103

radius of earth: 6.37 x 103 km
radius of Jupiter is not given.
I think error happens when Earth and Jupiter are aligned.


4. A varying electric current running through a coiled wire induces an emf across the coil. The voltage measured by a voltmeter across the terminals of the coil:
(a) is smaller or larger than the emf, depending on how fast the current changes.
(b) is larger than the emf.
(c) is equal to the emf.
(d) is smaller than the emf.

The answer is (d) but I thought it was (b).
My reasoning is that a voltage V is required to get the current running in the positive direction. Say the current is increasing in the positive direction. Then the magnetic flux is also increasing, so an induced emf in the negative direction is produced.
The measure voltage M = V - emf and since emf is relatively small, M > emf.

Also, please see https://www.physicsforums.com/showthread.php?t=574015

 

[anathema]

for more discussion on this topic.

Answer: The answer is (d) is correct. The voltage measured by the voltmeter across the terminals of the coil is smaller than the emf because some of the voltage is lost due to the internal resistance of the coil. This internal resistance causes a voltage drop, resulting in a smaller voltage measured by the voltmeter compared to the actual emf. This is known as the terminal voltage and is given by Vt = emf - IR, where I is the current and R is the internal resistance. Therefore, (d) is the correct answer.