solution to chapter 21 physics problems fundamentals of physics for free


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Suggested solution:
O = 3.0 cm, u = -14, f = -21 cm On simplification, v = -8.4 cm The image is erect, virtual and located 8.4 cm from the lens on the same side as the object. As the object is moved away from the lens, the virtual image moves towards the focus of the lens but never beyond. The image pro..
Numericals
Distance moved (s) = 10m Work done (W) = F x s 03. A work of 250J is done when a force of 10N is applied on an object. Calculate the distance through which the object moves. Solution:Work done (W) = 250 J Force applied (F) = 10 N Distance (s) = ? W = F x s Distance covered = 2..
Thermal-Chemical Effects of Currents Summary
. Therefore, for a mode of the substance deposited i.e., for m = M, where M is the relative atomic mass, the total charge flowing through the circuit is N A pe, where N A is the Avogadro's number. Thus, M = Z N A pe or Z = M/N A ep Where M/P is the equivalent mass of the substance. Thus Faraday's s..
Fundamental and Derived Units
Fundamental and Derived Units - The units of fundamental physical quantities are called fundamental units. They are length, mass and time. These units can neither be derived from one another nor can be resolved into any other units. They are independent of on..
PROBLEMS
1. The Earth's gravitational force causes an acceleration of 5 m/s 2 in a 1kg mass somewhere in space. How much will the acceleration of a 3kg mass be at the same place? Solution: The acceleration produced in any body due to the gravitational pull of the Earth does not depend on the mass ..
Problems
1. Calculate the force of attraction between Earth and the Sun. Solution: Given Mass of the Sun =10 3 2 kg Mass of the Earth = 6x10 2 4 kg Distance between the Sun and Earth = 150x10 9 m..
Suggested solution:
Intensity is proportional to the width of the slit. So, amplitude is proportional to the square root of the width of the slit. Here represent the widths of the two slits. ..
Suggested solution:
For an atom other than hydrogen, For lithium atom, Z = 3 So ionisation potential of lithium atom is the energy required to remove an electron (valence electron) from second orbit to infinity. ..
Suggested solution:
Now, = 3.28 x 10 1 5 Hz ..
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