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Question | Options | Answers | Marks | Category | Manage |
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Question | Options | Answers | Marks | Category | Manage |

Let S denote the set of all real values of X such that (x^2010+1) 1+x^2+x^4+…..+x^2008= 2010x^2009, then |
Option1: the number of elements in S is 2 Option2: the number of elements in S is 1 Option3: point (x,2) lies inside the parabola y =x^2-2x+2 Option4: Image of the point (x,2) in the line mirror y =x lies on x+y=4 |
option2 | 3 | Maths | Edit Delete |

If n∈N,f(n)=37^( n+2)+16^(n+1) +30 n then |
Option1: F(n)+1 is divisible by 3 Option2: F(n) is divisible by 3 Option3: F(n) is divisible by 7 Option4: F(4) is divisible by 5 |
option1 | 3 | Maths | Edit Delete |

If r,s and t be the roots of the equation ,8x^3+1001x+2008=0,then |
Option1: (r+s)^3+(s+t)^3+(t+r)^3=753 Option2: (r+s)^3+(s+t)^3+(t+r)^3= -753 Option3: 8(r+1)+(s+1)+(t+r)=999 Option4: 8(r+1)+(s+1)+(t+1)= - 999 |
option1 | 3 | Maths | Edit Delete |

The line x+y =1 meets x-axis at A and y-axis at B,P is the mid –point of AB; P1 is the foot of the perpendicular from P to OA; M1 is that of P1,on OP; P2 is that of M1 on OP; M2 is that of P2 on OP; P3 is that of M2 on OA; and so on. If Pn denotes the nth foot of the perpendicular on OA; then |
Option1: OPn=(1/2)^( n-1) Option2: OPn=(1/2)^ n Option3: OMn=(√2/2^(n+1)) Option4: OMn=(√2/2^n) |
option2 | 3 | Maths | Edit Delete |

If a, b, c are the side of a triangle ,then the value of a/(b+c-a)+b/(c+a-b)+c/(a+b-c) can be equal to |
Option1: 3 Option2: 4 Option3: 2 Option4: 3/2 |
option1 | 3 | Maths | Edit Delete |

Equation of circles through the origin ,making an intercept of √10 on the line y=2x +5/(√2), which subtends an angle of 45° at the origin is /are |
Option1: x^2+y^2-4x-2y=0 Option2: x^2+y^2-2x-4y=0 Option3: x^2+y^2+4x+2y=0 Option4: x^2+y^2+2x+4y=0 |
option2 | 3 | Maths | Edit Delete |

If (x-a)cos θ+ysin θ=x-acos ∅+ysin ∅=a and tan ( θ/2)-tan(∅2)=2b,then |
Option1: y^2=2ax-(1-b^2 )x^2 Option2: tan tan (θ/2)=(1/x)(y+bx) Option3: y^2=2bx-(1-a^2 )x^2 Option4: tan tan (∅/2)=(1/x)(y-bx) . |
option1 | 3 | Maths | Edit Delete |

If x^2+2xy+y^2=0 represent the equation of the straight lines through the origin which make an angle α with the straight line y+x=0, then |
Option1: sec sec2 α=h Option2: cos cosα=√(1+h)/2h Option3: 2 sin 2 sin α=√(1+h)/2h Option4: cot cot α=√(h+1)/(h-1) |
option1 | 3 | Maths | Edit Delete |

The number of ways of choosing triplets (x, y ,z) such that z>max(x, y) and x, y, z 1,2,…n,n+1is |
Option1: (n+1) C3+(n+2)C3 Option2: (1/6) n(n+1)(2n+1) Option3: 1^2+2^2+…+n^2 Option4: 2((n+2)C3)- (n+1)C2 |
option1 | 3 | Maths | Edit Delete |

The circle x^2 + y^2 + 6x – 24y + 72 = 0 and hyperbola x^2 – y^2 + 6x + 16y -46 = 0 intersect at four distinct points. These four points lies on a parabola then |
Option1: Coordinate of its focus is (-3,2) Option2: Coordinate of its focus is (3,0) Option3: Equation of directly is y = 0 Option4: Length of latus rectum is 4 |
option1 | 3 | Maths | Edit Delete |

A string is holding a solid block below the surface of the liquid as shown in figure. Now if the system is given an upward constant acceleration a, then as compared o previous state. |
Option1: Tension is string will be (1+a/g) times Option2: Tension in string will be (1-a/g) times Option3: Up thrust force on block become (1+a/g) times Option4: Up trust force on block become (1-a/g) times |
option1 | 3 | Physics | Edit Delete |

Two radio that are 250m apart emit sound waves of wavelength 100m. Point A is 400m from both station. Point B is 450 m from both station. Point C is 400m from one station and 450 m from the other. The radio station emit sound waves i phase. Which of the following statement is not true? |
Option1: There will constructive interference at A and B, and destructive interference at C. Option2: There will be destructive at A and B ,and constructive interference at C. Option3: There will be constructive interference at B and C ,and destructive interference at A. Option4: There will be destructive at A, B,C. |
option3 | 3 | Physics | Edit Delete |

A particle is projected with 0m/sec. From level ground on a vertical wall which is presented at a 5 m distance from the projection. Then which of following option’s is /are true.(Take g =10m/s2) |
Option1: Maximum possible height on wall at which particle can collide on wall is 3.75m Option2: For the case of maximum possible height, collision on wall ,the time of flight is 4 sec Option3: If collision between wall and particle is perfectly elastic(in the case of maximum possible height collision at wall) then particle will hit the ground at two meter from point of projection. Option4: For the particle to collide at maximum possible height on the wall the angle of projection from horizontal is θ=tan^-1(2) |
option4 | 3 | Physics | Edit Delete |

Consider a rope of mass 4m and length on a fixed pulley of radius R as shown in the figure. The rope is in equilibrium. Length of vertical hanging parts is shown in the figure. |
Option1: Torque of tension force about O on pulley is 4mg R Option2: Torque of normal force between rope and pulley about O is zero Option3: Torque of normal force between rope and pulley on pulley about O is mgR Option4: Torque of friction force between rope and pulley on pulley about O is zero |
option2 | 3 | Physics | Edit Delete |

A uniform body with circular cross –section released from rest on rough and fixed inclined plane shown in figure, then which of following is incorrect for the motion of object. |
Option1: If the body is hollow sphere and μ=.04then it will perform pure –rolling. Option2: If the body is hollow sphere and =.05 then ratio of rotational kinetic energy to the translational kinetic energy at bottom is 2/5 for given sphere. Option3: If the body is sphere and μ=0.1then the body will perform pure –rolling. Option4: If the body is solid sphere and μ=.05 then ratio of rotational kinetic energy to the translational kinetic energy at bottom is 2/3 for given sphere. |
option2 | 3 | Physics | Edit Delete |

1.2G OF iron pyrites (FeS2) is roasted to convert sulphur in FeSO2 to SO2with 80% efficiency. It is then oxidised to SO3 which is then absorbed in water to given H2SO4.If % yield of each of 2 reactions is either 60% or 30% then amount of H2SO4 can be: |
Option1: 6.82 mol Option2: 1.44 mol Option3: 3.62 mol Option4: 5.76 mol |
option2 | 3 | Chemistry | Edit Delete |

Number of electrons having l+m value equal to zero in 26Fe may be: |
Option1: 13 Option2: 1 Option3: 7 Option4: 12 |
option1 | 3 | Chemistry | Edit Delete |

Reaction is taking place at constant Temperature, Pressure &Volume, and then correct statement is/are: |
Option1: a + b = c + d Option2: Mavg. may increase or decrease depending upon limiting reagent. Option3: Vapour density of mixture will remain same throughout the course of reaction. Option4: Total moles will change with progress of reaction. |
option1 | 3 | Chemistry | Edit Delete |

For the dissociation A2B3(g)⇌2ABg+1/2B2(g) If ,M =Molecular mass of A2B3(g) D= Vapour density of A2B2(g) P°=Initial pressure of the correct statement(s): Then ,identify the correct statement (s): |
Option1: Equilibrium pressure can be expressed as 2P°D/M Option2: Equilibrium pressure can be expressed as P°M/2D Option3: Degree of dissociation can be expressed as M-2D/3D Option4: Increase in temperature will increase the magnitude of D |
option2 | 3 | Chemistry | Edit Delete |

For fixed amount of an ideal gas, three processes are carried out via AB,APB &AQB.Select correct options: |
Option1: ∆SAB does to depend upon the type of process. Option2: If AB path is a reversible iso thermal process , then ∆SAB=QAB/T, where QAB is heat involved in process AB and T is the temperature for the isothermal process. Option3: ∆SAB via AQB >∆SAB via APB. Option4: ∆SAB via AQB >∆SAB via APB. |
option2 | 3 | Chemistry | Edit Delete |