SCRA Physical Science |   1567

SCRA Physical Science

 


SCRA-Physical Sciences

 1. A block of glass having thickness 2 m is immersed 18 m deep in water. A ray of light is normally incident on the block after passing through the water. The time taken by the ray to cross the glass block from the moment of incidence on the water surface is  [Refractive index of glass is 3/2 and that of water is 4/3] 

(a) 9 * 10-8 s

(b) 8 * 10-8 s

(c) 2 * 10-8 s

(d) 10-8 s (ans)

Ans : Time taken by light ray to pass through the medium

= µx/c

= 3/2 * 2 = 10-8 s
   3 * 108

 2. The longest wavelength of Lyman series of hydrogen atom and the longest wavelength of Balmer series of the same spectrum bears the ratio

(a) 27 : 5

(b) 5 : 27 (ans)

(c) 4 : 1

(d) 1 : 4

Ans : The longest wavelength of Lyman series

l            = n2 (n + 1)2/(2n + 1)R
 max1

l            =  4 / 3R                (? n = 1)
 max1

The longest wavelength of Balmer series

l            = n2 (n + 1)2/(2n + 1)R
 max2

             = 36/5R                 (? n = 2)

 

∴       l             l         =   4 / 3R   
           max1 /   max2         36 / 5R

or       l             l         =   5    
           max1 /   max2         27

 Directions (Q. Nos. 3-4) Read the information carefully and answer the questions that follows

A cyclist rides along the circumference of a circular horizontal plane of radius R, the coefficient of friction being dependent only on the distance r from the centre of plane as

µ = µ0 (1 - r/R) where µ0 is  a constant

 3.  What is the radius of the circle with the centre at the point along which the cyclist can ride with maximum velocity ?

(a) R

(b) R/2 (ans)

(c) R/3

(d) R/4

 4.  What is the maximum velocity (with reference to above question) ?

(a) µ0 Rg / 4

(b) µ0 Rg / 2 (ans)

(c) õ0 Rg / 4

(d) None of these

 Solutions for Q. Nos. 3-4

µ = µ0 (1 - r/R)

µ = µ0 (R - r/R)

µR = µ0 (R - r)

At  r = R/2, the cyclist can ride with maximum velocity

µRg = µ0 (R - R/2) g = µ0 R/2 g

µRg ≥  µ0Rg / 2

∴ The maximum velocity of cyclist =  µ0Rg / 2

5.  A particle moving with simple harmonic motion has an amplitude A. Its distance from the centre, where the velocity is half of the maximum value, will be

(a) ± A/√2

(b) ± A/√3

(c) ± √3 A/2 (ans)

(d) ± A/2

Ans :     v        =    wA
             max 

∴        v        =    wA/2 = w√A2 - y2

or     A2 - y2   =  A2/4

or           y2   = 3A2/4

or              y = ± √3A/2

 

6. A plane mirror is placed vertically 15 cm away from the image (I) formed by a concave mirror of focal length 20 cm when a pencil (O) is placed at a distance of 30 cm from the latter. The position of the image (I') from the concave mirror formed by the plane mirror, is

(a) 30 cm

(b) 60 cm

(c) 75 cm

(d) 90 cm  (ans)

Ans :

1/f = 1/v + 1/u

- 1/20 = 1/v - 1/30

1/v =  1/20 + 1/30

1/v = -3 + 2 / 60 = - 1/60

v = - 60 cm

We can see in the figure that the position of the image (I') from the concave mirror formed by the plane mirror is 90 cm from the concave mirror.

 

7. A near-sighted man can clearly see up to a distance of 1.5 m. What is the power of the lens of the spectacles necessary for the remedy of the defect ?

(a) + 0.67 D

(b) - 0.67 D  (ans)

(c) + 1.5 D

(d) - 1.5 D

Ans :     u¥

and v =  1.5 m

1/f = 1/v + 1/u

1/f = 1/1.5 - 1/¥

1/f = - 1.5

∴ power = 1/-1.5 = -0.666

= -0.67 D

 

8. Two radioactive nuclei have half lives 15 min and 20 min respectively. Initially, they have equal number of nuclei. What is the ratio of their remaining nuclei after 1 h ?

(a) 1 : 2 (ans)

(b) 2 : 1

(c) 3 : 4

(d) 4 : 3

Ans :  For first radioactive nuclei,

N1/N0 = (1/2)t/T

N1/N0 = (1/2)60/15

N1/N0 = (1/2)4

N1/N0 =  1/16

For second radioactive nuclei,

N2/N0 =  (1/2)60/20

N2/N0 =  (1/2)3

N2/N0 = 1/8

 

9. Consider the following statement :

I. The focal length of a lens is dependent on the wavelength of light

II. The refractive index of the material of the lens is not same for all wavelengths of light

Which one of the following is correct in respect of the above statements ?

(a) Both the statement are true and Statement II is the correct explanation of Statement I

(b) Both the statements are true but Statement II is not the correct explanation of Statement I (ans)

(c) Statement I is true, but Statement II is false

(d) Statement I is false, but Statement II is true

Ans :  Focal length of a lens

1/f = (µ-1)[1/R1 - 1/R2]

where µ is the refractive index of material By Cauchy formula,

µ = A + B/l2 + C/l4 + ....

Therefore, refractive index of the material of the lens is dependent on the wavelength of light. So, focal length of a lens depends upon the wavelength of light. So, option (b) is correct.

 

10. The sum of the individual masses of a 3Li7 nucleus is 0.042 u more than the actual mass of the  3Li7 nucleus. What is the binding energy per nucleon of the nucleus ?

[1 u = 931 MeV]

(a) 13.034 MeV

(b) -13.034 MeV

(c) 5.586 MeV (ans)

(d) -5.586 MeV

Ans :  Given, ?m = 0.042

Binding energy per nucleon of the 3Li7 nucleus

= ?m * 931/A

= 0.042 * 931
            7

= 5.586 MeV

 

11. Which of the following compounds exhibit optical activity ?

 

Select the correct answer using the codes given below.

Codes

(a) I and II

(b) I and IV

(c) II and III  (ans)

(d) III and IV

Ans : Compounds containing chiral carbon, the carbon all the four valancies of which are statisfied by four different groups, are optically active.

 *
(C = Chiral carbon atom)

Thus, II and III exhibit optical activity

 

12. When propylene reacts with hydrogen bromide in presence of peroxide, the product formed is

(a) n-propyl alcohol

(b) propylene peroxide

(c) n-propyl bromide  (ans)

(d) iso-propyl bromide

 

CH3CH2CH2Br

n - propyl bromide

(In the presece of peroxide, the addition takes place according to Anti-Markownikoff's rule. This is also called Kharash effect.)

Note : According to Anti-Markownikoff's law, the negative part of the adduct goes to that carbon atom which contains more number of hydrogen atoms.

 

13. Consider the following statements :

I. p-nitrostyrene easily undergoes nucleophilic addition reaction with diethylamine but styrene can not.

II. The terminal methylene group in p-nitrostyrene carries a negative charge and behaves as a nucleophile.

Which one of the following is correct in respect of the above two Statements ?

(a) Both the statements are true and Statement II is the correct explanation of Statement I

(b) Both the Statements are true but Statement II is not the correct explanation of Statement I

(c) Statement I is true but Statement II is false  (ans)

(d) Statement I is false but Statement II is true

Ans : The resonating structures of p-nitrostyrene are as
 


 

(Because - No2 is a strong deactivating group and thus, decreases electron density and facilitate nucleophilic addition reaction.)

In case of styrene, is weakly activating and thus, increases electron density. Thus, styrene does not show nucleophilic addition reaction with diethylamine

 

14. Consider the following reaction :

What is the major product [X] of the reaction ?

(b)

 

(c)

 

(d)

 (ans)

 

Ans :  

 

 

 

 

 

15. The enzyme pepsine hydrolyses

(a) glucose to ethyl alcohol

(b) fats to fatty acids

(c) polysaccharides to monosaccharides

(d) proteins to amino acids  (ans)

Ans : Proteins

 Amino acids

 

Similar Categories
more 
ONLINE TESTS
feedback