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General

General

Easy

Question

The first meiotic division during oogenesis is completed at the stage of

Primary oocyte within primary follicle
Primary oocyte within secondary follicle
Primary oocyte within tertiary follicle
Secondary oocyte within tertiary follicle

The correct answer is: Primary oocyte within tertiary follicle

Related Questions to study

If $fraction numerator left parenthesis x squared plus 3 right parenthesis left parenthesis x minus 1 right parenthesis to the power of 4 over denominator left parenthesis x plus 1 right parenthesis end fraction greater than 0 space t h e n space x element of$

If $fraction numerator left parenthesis x squared plus 3 right parenthesis left parenthesis x minus 1 right parenthesis to the power of 4 over denominator left parenthesis x plus 1 right parenthesis end fraction greater than 0 space t h e n space x element of$

Assertion : In any ABC, minimum value of $fraction numerator r subscript 1 end subscript plus r subscript 2 end subscript plus r subscript 3 end subscript over denominator r end fraction$ is 9.
Reason : A.M.  G.M.

Assertion : In any ABC, minimum value of $fraction numerator r subscript 1 end subscript plus r subscript 2 end subscript plus r subscript 3 end subscript over denominator r end fraction$ is 9.
Reason : A.M.  G.M.

Assertion: The orthocentre of the given triangle is coincident with the in-centre of the pedal triangle of the given triangle.
Reason : Pedal triangle is the ex-central triangle of the given triangle.

Assertion: The orthocentre of the given triangle is coincident with the in-centre of the pedal triangle of the given triangle.
Reason : Pedal triangle is the ex-central triangle of the given triangle.

parallel

A large number of primary follicles degenerate during the phase from birth to puberty. Therefore at puberty each ovary has about

A large number of primary follicles degenerate during the phase from birth to puberty. Therefore at puberty each ovary has about

Assertion : The side of regular hexagon is 5 cm whose radius of inscribed circle is 5cm.
Reason : The radius of inscribed circle of a regular polygon of side a is $fraction numerator a over denominator 2 end fraction cot invisible function application open parentheses fraction numerator pi over denominator n end fraction close parentheses$ .

Assertion : The side of regular hexagon is 5 cm whose radius of inscribed circle is 5cm.
Reason : The radius of inscribed circle of a regular polygon of side a is $fraction numerator a over denominator 2 end fraction cot invisible function application open parentheses fraction numerator pi over denominator n end fraction close parentheses$ .

Assertion : In a ABC, $fraction numerator a cos invisible function application A plus b cos invisible function application B plus c cos invisible function application C over denominator a plus b plus c end fraction$ is equal to $fraction numerator r over denominator R end fraction$
Reason : In equilateral triangle the ratio between In-radius and circum-radius is 1 : 2.

Assertion : In a ABC, $fraction numerator a cos invisible function application A plus b cos invisible function application B plus c cos invisible function application C over denominator a plus b plus c end fraction$ is equal to $fraction numerator r over denominator R end fraction$
Reason : In equilateral triangle the ratio between In-radius and circum-radius is 1 : 2.

parallel

Assertion : In any triangle ABC, $fraction numerator 1 over denominator a b end fraction plus fraction numerator 1 over denominator b c end fraction plus fraction numerator 1 over denominator c a end fraction equals fraction numerator 1 over denominator 2 r R end fraction$ , where r is in radius and R is circum radius.
Reason : R  2r.

Assertion : In any triangle ABC, $fraction numerator 1 over denominator a b end fraction plus fraction numerator 1 over denominator b c end fraction plus fraction numerator 1 over denominator c a end fraction equals fraction numerator 1 over denominator 2 r R end fraction$ , where r is in radius and R is circum radius.
Reason : R  2r.

In any equilateral , three circles of radii one are touching to the sides given as in the figure then area of the 

In any equilateral , three circles of radii one are touching to the sides given as in the figure then area of the 
Maths-General

If the sides a, b, c of a triangle are such that a : b : c : : 1 : $square root of 3$ : 2, then the A : B : C is -

If the sides a, b, c of a triangle are such that a : b : c : : 1 : $square root of 3$ : 2, then the A : B : C is -

parallel

If the angles of a triangle are in ratio 4 : 1: 1 then the ratio of the longest side and perimeter of triangle is -

If the angles of a triangle are in ratio 4 : 1: 1 then the ratio of the longest side and perimeter of triangle is -

Which of the following pieces of data does NOT uniquely determine an acute angled triangle ABC (R being the radius of the circumcircle) -

Which of the following pieces of data does NOT uniquely determine an acute angled triangle ABC (R being the radius of the circumcircle) -

In a triangle ABC, let C = $fraction numerator pi over denominator 2 end fraction$ . If r is the in radius and R is the circumradius of the triangle, then 2(r + R) is equal to -

In a triangle ABC, let C = $fraction numerator pi over denominator 2 end fraction$ . If r is the in radius and R is the circumradius of the triangle, then 2(r + R) is equal to -

parallel

The solution set of the equation $fraction numerator 1 over denominator left parenthesis x plus 3 right parenthesis end fraction less or equal than negative 2 space i s$

The solution set of the equation $fraction numerator 1 over denominator left parenthesis x plus 3 right parenthesis end fraction less or equal than negative 2 space i s$

Solution of $left parenthesis 5 x minus 1 right parenthesis less than left parenthesis x plus 1 right parenthesis squared less than left parenthesis 7 x minus 3 right parenthesis$ is

Solution of $left parenthesis 5 x minus 1 right parenthesis less than left parenthesis x plus 1 right parenthesis squared less than left parenthesis 7 x minus 3 right parenthesis$ is

Let A0 A1 A2 A3 A4 A5 be a regular hexagon inscribed in a circle of unit radius. Then the product of the lengths of the line segments A0A1, A0A2 , and A0A4 is -

Let A0 A1 A2 A3 A4 A5 be a regular hexagon inscribed in a circle of unit radius. Then the product of the lengths of the line segments A0A1, A0A2 , and A0A4 is -

parallel

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