चित्र `1.27` में विधुत क्षेत्र अवयव `E_(x)= alpha x^(1//2), E_(y)=O` है, जिसमें `alpha =800 N//C m^(1//2)` है । (a) घन से गुजरने वाला फ्लक्स , तथा (b) घन के भीतर आवेश परिकलित कीजिये । `alpha =0.1 m` मानिये ।

The electric field components in Fig. are E_(x)=ax^(1//2), E_(y)=E_(z)=0 , in which alpha=800N//C m^(1//2) . Calculate the flux through the cube

The electric field components in Figure are E_(x)=alpha x^(1/2), E_y=E_(z)=0 in which alpha=800 N//Cm^(1/2) . Calculate (a) the flux through the cube, and (b) the charge within the cube. Assume that a=0.1m.

The electric components in the figure are E_(x)=alphax^(1//2) ,E_(y)=0,E_(z)=0 where alpha=800N//m^(2) if a=0.1m is the side of cube then the charge with in the cube is

The electric components in the figure are E_(x)=alphax^(1//2) ,E_(y)=0,E_(z)=0 where alpha=800N//m^(2) if a=0.1m is the side of cube then the charge with in the cube is

If int_(0)^(1) x e^(x^(2) ) dx=alpha int_(0)^(1) e^(x^(2)) dx , then

If 2 cos alpha = x + (1)/(x) and 2 cos beta = y + (1)/(y) , show that x^(m) y^(n) + (1)/(x^(m)y^(n)) = 2 cos (m alpha + n beta)

Let f (x) = log _e (sinx ), ( 0 lt x lt pi ) and g(x) = sin ^(-1) (e ^(-x)), (x ge 0) . If alpha is a positive real number such that a = ( fog)' ( alpha ) and b = (fog ) ( alpha ) , then (A) aalpha ^(2) - b alpha - a = 0 (B) a alpha ^(2) - b alpha - a = 1 (C) a alpha ^(2) + b alpha - a = - 2 alpha ^(2) (D) a alpha ^(2) + b alpha + a = 0