Monday, August 23, 2021

Properties Of Bulk Matter: Physics

  • Answers are in BOLD Option name.


Q.1. An iron bar of length l and cross-section A is heated from 0 degree C space t o space 100 degree C. If the rod is so held that it is not permitted to expand or bend, the force developed in it is:

A) directly proportional to length

B) inversely proportional to length

C) independent of length

D) inversely proportional to area of cross-section

Explanation-

Change in length increment l equals alpha divided by increment T


Stress = F over A space a n d space S t r a i n equals fraction numerator increment l over denominator l end fraction

N o w space Y equals F over A cross times fraction numerator l over denominator increment l end fraction equals F over A cross times fraction numerator l over denominator alpha l increment T end fraction o r space F equals Y space alpha space A space increment space T.

Clearly F does not depend upon length.


Q.2. The radius of a spherical soap bubble is 0.5 mm. If the surface tension of the soap solution be 30 dyne/ cm, the excess pressure is :

A) 1500 space d y n e divided by c m squared

B) 1000 space d y n e divided by c m squared

C) 2400 space d y n e divided by c m squared

D) 1200 space d y n e divided by c m squared

Explanation-

For soap bubble, excess pressure is 

        P subscript i minus P subscript 0 equals fraction numerator 4 T over denominator r end fraction space space space space space space space space space space space space space space space space space space space equals fraction numerator 4 cross times 30 over denominator 0.5 cross times 10 to the power of negative 1 end exponent end fraction equals 2400 space d y n e divided by c m squared 


Q.3. A thick rope of rubber of density 1.5 kg/m cubed and Y equals 5 cross times 10 to the power of 6 space N space m to the power of negative 2 end exponent metres in length when hung vertically will increase in length by :

(g=10 m s to the power of negative 2 end exponent )

A) 9.6 cross times 10 to the power of negative 5 end exponent space m

B) 9.6 space m

C) 19.2 cross times 10 to the power of negative 3 end exponent space m

D) 19.2 cross times 10 to the power of negative 5 end exponent space m 

Explanation-

As the weight of wire is at center of gravity

therefore space O n l y space h a l f space t h e space l e n g t h space o f space r o p e space g e t s space e x t e n d e d. space space space space space space space space space space space Y equals F over A. fraction numerator open parentheses L divided by 2 close parentheses over denominator increment l end fraction. therefore space space space increment l equals fraction numerator M g over denominator A Y end fraction. L over 2 equals fraction numerator A L p g. h over denominator 2 A y end fraction equals fraction numerator L squared p g over denominator 2 Y end fraction. space space space space space space space space space space space space equals fraction numerator 64 cross times 1.5 cross times 10 over denominator 2 cross times 5 cross times 10 to the power of 6 end fraction equals 9.6 cross times 10 to the power of negative 5 end exponent space m 

Q.4. The density of ice is x g c m to the power of negative 3 end exponent and that of water is y g c m to the power of negative 3 end exponent. What is the change in volume when m gram of ice melts ?

A) m ( y - x ) c.c

B) fraction numerator y minus x over denominator m end fraction space c. c

C) m space x y space left parenthesis x minus y right parenthesis space c. c

D) m open parentheses 1 over y minus 1 over x close parentheses space c. c 

Explanation-

Vol. of m gram water formed equals m over y space c. c

Change in vol. on melting open parentheses m over y minus m over x close parentheses

                      equals m open parentheses 1 over y minus 1 over x close parentheses space c. c 


Q.5. Water rises to a height of 10 cm in capillary tube and mercuy falls to a depth of 3.42 cm in the same tube. If the density of mercury is 13.6 g  c m to the power of negative 3 end exponent and angle of contact is 135 degree, the ratio of the surface tensions for water and mercury is :

A) 1 : 5.57

B) 1 : 3.57

C) 1 : 6.57

D) 1 : 1.57

Explanation-

T subscript omega over T subscript m equals fraction numerator h subscript omega d subscript omega over denominator cos space theta subscript omega end fraction cross times fraction numerator cos space theta subscript m over denominator h subscript m cross times d subscript m end fraction equals fraction numerator 10 cross times 1 cross times cos space 135 degree over denominator 1 cross times 3.42 cross times 13.6 end fraction space space space space space space space equals fraction numerator 10 cross times 1 over denominator 3.42 cross times 1.414 cross times 13.6 end fraction equals fraction numerator 1 over denominator 6.57 end fraction 


Q.6. The isothermal bulk modulus of an ideal gas at a pressure P is:

A) P

B) y P

C) P/2

D) P/y

Explanation-

 For isothermal process PV + constant 

Differentiating both sides,

p d V space plus space V d p space equals space 0 space space space space space space space space space space space space space space space space space space space space P d v space equals space minus space V space d p space space rightwards double arrow space space space space space space P equals negative fraction numerator V d p over denominator d V end fraction. o r space space space space space space P equals negative fraction numerator d P over denominator open parentheses d V divided by V close parentheses end fraction o r space sin c e space b u l k space m o d u l u s comma space K equals negative fraction numerator d P over denominator d V divided by V end fraction. therefore space space space space space space space space space space space space space space space space space space K plus P 


Q.7. The reading of a pressure meter attached with a closed water pipe is 3.5 cross times 10 to the power of 5 space N divided by m squared. On opening the valve, the reading of the pressure meter is reduced to 3 cross times 10 to the power of 5 space N divided by m squared. Calculate the speed of water flowing in the pipe.

A) 10 m/s

B) 20 m/s

C) 30 m/s

D) 40 m/s.

Explanation-

According to Bernoulli's theorem,

  P subscript 1 over p plus 1 half space v subscript 1 squared equals P subscript 2 over p plus v subscript 2 squared fraction numerator 3.5 cross times 10 to the power of 5 over denominator 10 cubed end fraction plus 0 equals fraction numerator 3 cross times 10 to the power of 5 over denominator 10 cubed end fraction plus 1 half v subscript 2 squared 350 minus 300 equals 1 half v subscript 2 squared space space space space space space space space space v subscript 2 squared equals 50 cross times 2 equals 100 space space space space space space space space space v subscript 2 squared equals 10 space m divided by s. 


Q.8. A circular tub filled with liquid is uniformly tapering with base area 10 to the power of negative 2 end exponent space m squared and top area 2 cross times 10 to the power of negative 3 end exponent space m squared. Its depth is 0.4 m as shown in fig. What is the thrust on its base ? open parentheses g equals 10 space m s to the power of negative 2 end exponent comma space p equals 900 space k g space m to the power of negative 3 end exponent close parentheses space. 

A) 3.6 N

B) 7.2 N

C) 9 N

D) 14.4 N

Explanation-

Here thrust at bottom equals space h d g space cross times space A

                     F equals 0.4 cross times 900 cross times 10 cross times 10 to the power of negative 3 end exponent space space space equals 3.6 space N 


Q.9. An incompressible liquid flows through a horizontal tube bifurcating at the end as shown

The areas are expressed in m squared and velocity in m space s to the power of negative 1 end exponent. What is the value of v subscript 0 ?

A) 3 m space s to the power of negative 1 end exponent

B) 1.5 m space s to the power of negative 1 end exponent

C) 1 m space s to the power of negative 1 end exponent

D) 1.25 m space s to the power of negative 1 end exponent

Explanation-

Here a v subscript 0 equals a subscript 1 v subscript 1 plus a subscript 2 v subscript 2 cross times 3 equals a cross times 1.5 plus 1.5 a cross times v subscript 2 o r space space space space space space space space v subscript 2 equals 1 space m s to the power of negative 1 end exponent 


Q.10.  A log of wood floats in water with 1/5th of its volume above the surface of water. The density of wood is :

A) 0.8 cross times 10 cubed space k g space m to the power of negative 3 end exponent

B) 8 cross times 10 cubed space k g space m to the power of negative 3 end exponent

C) 0.08 cross times 10 cubed space k g space m to the power of negative 3 end exponent

D) 10 cubed space k g space m to the power of negative 3 end exponent 

Explanation-

Here the wt. of the body = Wt. of water displaced by immersed portion of the body

        space space space space space space space space space space space space space space space space V space cross times p g equals 4 over 5 space V. space d. space g therefore space space space space space space space space space space space space space space space space space space space p equals 4 over 5. space d equals 08 cross times 10 cubed space k g space m to the power of negative 3 end exponent. 



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