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amulye (84)

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a metal piece of mass 160g lies in equilibrium inside a glass of water d piece touches d bottom of d glass at a small no. of points if d density of d metal is 8000kg/m^3find d normal force exerted by d bottom of d glass on d metal piece.
 
if error in d measurement of mass  is 0.8%& in vol. it is 0.4% den error in d density is ( ans:1.2% )
 
A hollow spherical body of inner & outer radii 6cm &8cm resply floats half submerged in water find d density of d material of sphere
 
a reflecting surface is represented  by d equation
 
Y=2L/  Sin(x/L), where 0 xL
 
A ray of light travelling horizantally  becomes vertical after reflection  with d surface d co ordites of d point where dis ray is incident is [ans : L/3,3L/5]
 
 
water leaks out from an popen tank thru a hole of area 2mm^2 in d bottom suppose water is filled upto a height of 80cm & d area of cross section of d tank is 0.4m^2 d pressure at d open surface & at d hole r equal to atmospheric pressure neglect d small velocity of water near d open surface in d tank
a)find d initial speed of water coming out of d hole b) find d sped of water coming out when 1/2 of d water has leaked out c)find d vol of water leaked out during a time interval dt after d height remained is h dus find d decrease in d height dh in terms of h &dt from d result of part c find d time reqd for d 1/2 of water to leak out
 
 
 
    
amulye (84)

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edit its actually L/pi rt3
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amulye (84)

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uhoooooooooo dont anybody ha guts to do dis plz help me
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allamraju (633)

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2)(d/d)X100=(m/m+v/v)X100=0.8+0.4=1.2%
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Shivam18 (15)

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At equilibrium ,

Mg = N +V(rho-sigma)g

therefore , N = 0.2 newtons
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amulye (84)

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plz tell d ans frnds
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