To solve this problem it is necessary to apply the concepts related to mutual inductance in a solenoid.
This definition is described in the following equation as,

Where,
permeability of free space
Number of turns in solenoid 1
Number of turns in solenoid 2
Cross sectional area of solenoid
l = Length of the solenoid
Part A )
Our values are given as,





Substituting,



PART B) Considering that many of the variables remain unchanged in the second solenoid, such as the increase in the radius or magnetic field, we can conclude that mutual inducantia will appear the same.
Answer:
Explanation:
Work done in lifting the weight once = mgh
= 20 x 9.8 x (1.9+1.7)
= 705.6 J
= 705.6 / 4.2 calorie
= 168 cals
Total energy to be spent = 600 x 10³ cals
No of times weight is required to be lifted
= 600 x 10³ / 168
= 3.57 x 10³ times
Total time to be taken = 2 x 3.57 x 10³
= 7.14 x 10³ s
=119 minutes .
Answer:
a) m_v = m_s ((
)² - 1) , b) m_v = 1.07 10⁻¹⁴ g
Explanation:
a) The angular velocity of a simple harmonic motion is
w² = k / m
where k is the spring constant and m is the mass of the oscillator
let's apply this expression to our case,
silicon only
w₉² =
k = w₀² m_s
silicon with virus
w² =
k = w² (m_v + m_s)
in the two expressions the constant k is the same and q as the one property of the silicon bar, let us equal
w₀² m_s = w² (m_v + m_s)
m_v = (
)² m_s - m_s
m_v = m_s ((
)² - 1)
b) let's calculate
m_v = 2.13 10⁻¹⁶ [(
)² - 1)]
m_v = 1.07 10⁻¹⁴ g