Miscellaneous continued

ADDED Q3)
Q1)
the plane of a dip circle is set in the geographical meridian and the apparent dip is θ1. It is then set in a vertical plane perpendicular to the geographic meridian. The apparent dip angle is θ2. The declination θ at the place is

A) θ = tan-1( tanθ1.tanθ2) B) θ = tan-1( tanθ1+tanθ2) C) \theta = tan^{-1}\left(\frac{tan\theta _{1}}{tan\theta _{2}} \right)
D) θ = tan-1( tanθ1 - tanθ2)

Q2)
two persons , A and B, stand at a distance x away frm each other in an open field. Wind blows frm A to B with a velocity ' w '.
i) A has a drum and sound travels frm A to B in t1 seconds.
ii) Now B has the drum and sound travels frm B to A in t2 seconds.

the velocity of sound in still air is :

A) 2x(t1+t2) B) \frac{x}{2}\left[\frac{1}{t_{1}}+\frac{1}{t_{2}} \right] C) \frac{wx}{2}\left[\frac{1}{t_{1}}+\frac{1}{t_{2}} \right] D) 2wx(t1+t2)

18 Answers

1
ABHI ·

i did it using refractive index....dnt knw hw akshay bhaiya did it

1
aposlil ·

Akshay's correct. Answer is D). Doppler effect is applicable in this case.

39
Pritish Chakraborty ·

Yaar even I think Akshay is right...a star moving away provides a red shift. Doppler's effect in light has to be applied somehow..
Define "special" abhirup. lol..

Q1. All I know is that there is a relation
cot²∂1 + cot²∂2 = cot²∂
Now how you're going to get the answer from that, I have no idea...

1
ABHI ·

so d shud b answer, now v hv 2 think of a valid reason
my reason is using refractive index, refractive index decreasin so wavelngth must increase.
n akshay bhaiya sayin abt doppler effect, as it is goin away, its frequency decreases so wavelenth must increase..but as far as i read, doppler effect is not valid for spacial things...
tabh valid reason kya hoga?

1
Akshay Pamnani ·

@ Abhirup
maybe u r correct,but then wat is the use of the fact that star is going away??

1
ABHI ·

i jst want 2 confirm my way of thinking frm akshay bhaiya first....n akshay bhaiya, doppler effect is not applicable for cosmic particles...so how r u puttin it 4 the star thing?

1
Akshay Pamnani ·

@ Abhirup
ya but then wat is the use of the fact that the star is going away
I am sure Doppler's effect has to be used here

1
Manmay kumar Mohanty ·

can u explain a bit more ABHIRUP

1
ABHI ·

cant v thing of using refractive index...n laboratory is certainly denser than the outer space....
so refractive index is lower in the outer space...so the wavelength must increase

1
Avinav Prakash ·

velo of air from A to B=v+w
so t1=xv+w

velo of air from B to A=v-w

t2=xv-w

use this reln to gwt b) as ans

1
Akshay Pamnani ·

Doppler's effect
appearent frequency would decrease and wavelength increase
But is the Doppler's effect applicable to Hydrogen Spectrum??

1
Manmay kumar Mohanty ·

can u explain how...............
I culdn't thnk of anythng when i saw this problem [2]

1
Akshay Pamnani ·

D hoga bhai

1
Manmay kumar Mohanty ·

kyun nahin clear to hai........

1
Akshay Pamnani ·

@ Manmay
mujhe dikh nahi raha Q3

1
Manmay kumar Mohanty ·

ANYONE WITH QUESTION 3 ) ???

1
Manmay kumar Mohanty ·

Q3)
The four line Balmer series spectrum shown below is emitted by a hydrogen sample in a laboratory. A star moving away frm Earth also emits a hydrogen spectrum.

WHICH spectrum might be observed on Earth for this star ?

IN abv options wavelngth ( x 10 -7 m)

1
Manmay kumar Mohanty ·

yup that's correct avinav [1]

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