The energy E of a particle at position x at time t is given by E=a/(t(b+x^(2)) Where a and b are constants. The dimensional formula of a is

The potential energy of a particle varies with distance x as U=(Ax^(1//2))/(x^(2)+B) where a and B are constants. The dimensional formula for AXB is :

The position x of a particle at time t is given by : x=(v_(0))/(a)(1-e^(-at)) where v_(0) is a constant and a>0. The dimensional formula of v_(0) and a is :

The potential energy of a particle varies with distance 'x' as U=(Ax^(1/2))/(x^(2)+B),where A and B are constants.The dimensional formula for A times B is [M^(a)L^(b)T^(c)] then a=

The position of a particle as a function of time t, is given by x(t)=at+bt^(2)-ct^(3) where a,b and c are constants. When the particle attains zero acceleration, then its velocity will be :

The velocity v of a particle at time t is given by v = a t + \frac { b } { t + c } where a, b and c are constant. The dimensions of a, b and c respectively are

If the velocity of a particle "v" as a function of time "t" is given by the equation,v=a(1-e^(-bt) ,where a and b are constants,then the dimension of the quantity a^2*b^3 will be

A particle of mass m is located in a region where its potential energy [U(x)] depends on the position x as potential Energy [U(x)]=(a)/(x^2)-(b)/(x) here a and b are positive constants… (i) Write dimensional formula of a and b (ii) If the time perios of oscillation which is calculated from above formula is stated by a student as T=4piasqrt((ma)/(b^2)) , Check whether his answer is dimensionally correct.