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A new unit of length is chosen such that the speed of light in vacuum is unity. What
is the distance between the Sun and the Earth in terms of the new unit if light takes
8 min and 20 s to cover this distance ?


Distance space between space sun space and space earth space is comma space

speed space of space light space cross times space time space taken space by space light space to space cover space the space distance space

Given comma space

Speed space of space light space equals space 1 space unit

Time space taken comma space straight t space equals space 8 space min space 20 space sec space equals space 500 space straight s space

So comma space

Distance space between space the space sun space and space the space earth space equals space 1 cross times space 500

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 500 space units
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Answer the following : 
(a)You are given a thread and a metre scale. How will you estimate the diameter of
the thread ?


The diameter of the thread is so small that it cannot be measured using a metre scale. 

Wind a number of turns of threads on the metre scale so that the turns are closely touching one another.

Measure the length (l) of the windings on the scale which contains n number of turns.

Error converting from MathML to accessible text.

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Which of the following is the most precise device for measuring length : 
(a) a vernier callipers with 20 divisions on the sliding scale
b) a screw gauge of pitch 1 mm and 100 divisions on the circular scale

(c) an optical instrument that can measure length to within a wavelength of light.



A device with minimum count is the most suitable to measure length.

a) 

straight L. straight C space of space vernier space calliper space equals space 1 space straight S. straight D space minus space 1 space straight V. straight D

space space space space space space space space space space space space space space space space space equals space 1 space minus space 9 over 10 space equals space 1 over 10 space equals space 0.01 space cm  

b) 

straight L. straight C space of space screw space gauge space equals space fraction numerator Pitch over denominator Number space of space divisions end fraction space

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 1 over 1000 space equals space 0.001 space cm space

straight c right parenthesis

straight L. straight C space of space an space optical space device space equals space Wavelength space of space light space tilde 10 to the power of negative 5 end exponent space cm space

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 0.00001 space cm 

Hence, it can be inferred that an optical instrument is the most suitable device to measure length.




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A student measures the thickness of a human hair by looking at it through a microscope of magnification 100. He makes 20 observations and finds that the average width of the hair in the field of view of the microscope is 3.5 mm. What is the estimate on the thickness of hair ?


Magnification of the microscope = 100 

Average width of hair in the field of view of the microscope = 3.5 mm 

Therefore, actual thinckness of the hair = 35 over 100 space equals space 0.035 space mm 

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Explain this statement clearly : 
“To call a dimensional quantity ‘large’ or ‘small’ is meaningless without specifying a
standard for comparison”. In view of this, reframe the following statements wherever
necessary : 
(a) atoms are very small objects 
(b) a jet plane moves with great speed 
(c) the mass of Jupiter is very large 
(d) the air inside this room contains a large number of molecules 
(e) a proton is much more massive than an electron 
(f) the speed of sound is much smaller than the speed of light

The given statement is true because a dimensionless quantity may be large or small in comparison to some standard reference. For example, the coefficient of friction is dimensionless. The coefficient of sliding friction is greater than the coefficient of rolling friction, but less than static friction. 

a) An atom is a very small object in comparison to a soccer ball.

b) A jet plane moves with a speed greater than that of a bicycle.

c)  Mass of Jupiter is very large as compared to the mass of a cricket ball.

d)  The air inside this room contains a large number of molecules as compared to that present in a geometry box.

e) A proton is more massive than an electron.

f) Speed of sound is less than the speed of light.

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