Practice Problems for IJSO Stage-1Full description
Correction in Official NSEJS (IJSO) Stage-I Answer Key Paper Code : 514 MATHEMATICS MATHEMATICS : 46.
Around a lawn which is of semicircular shape a pavement of uniform width is laid. If he ratio of the ara of the lawn to the area of the pavement is 25:24 , thenthe ratio of the outer and inner perimeters of hte pavement is a)
7 5
b)
6 5
c)
5 4
d)
5 2 6
Official Solution :(b)
Explaination from Reso-PCCP side :
46.
"r 2 Area of lawn 25 ! 2 2 2 ! Area of pavement "r1 # "r2 24 r
Let ABC be a triangle in which AB = AC. Let D be a point on AC such that BD bisects angle B. Value of the ratio
AB is between BC
a) 1.0 and 1.5
b) 1.5 and 2.0
Official Solution :(b)
Explaination from Reso-PCCP side : All options are possible
c) 2.0 and 2.5
d) 2.5 and 3.0
62.
In the xy-plane let A be the point (5.0) and L be the line y =
x . The number of poins % on the line L such 3
that triangle OAP is isosceles is (O being the origin) a) 1
b) 2
Official Solution :(c)
Explaination from Reso-PCCP side :
62.
P lies an y =
x h $ h = 3K = 50 K = 3 3
P (3K,K) Now 'OAP is isoscles if (i) OP = PA (ii) PA = OA (iii) OA = OP Coee (i) OP = PA $ OP2 = PA PA2 (3K – 0)2 + (K – 0)2 = (3K – 5)2 + (K – 0)2 9k2 + k2 = 9k2 – 30k + 25 + k 2 $ 30K = 25 K=
- 5 5 * 25 5 $ K = so P+ 2 , 6 ( , ) 30 6
Case (ii) PA = OA PA2 = OA2 (3K – 5)2 + (K – O)2 = 52 $ 9K2 – 30 K + 25 + K2 = 25 $ 10K2 – 30K = 0 $ 10K ( K – 3) = 0 K = 0 / K = 3 P (0 (0,0) ,0) {no {nott po possi ssible} ble} P (9, (9,3) 3) Case (iii) OA2 = OP2 52 = (3k – O)2 + (K – O)2 25 = 9k2 + k2 $ 10k2 = 25 5 K2 = 2 K=±
5 2
- 5 5 * + ( P +3 2 , 2 ( , )
c) 3
d) infinitely many
- 5 5 *( + – P + – 3 2 , – 2 )( , total 4 points are there 70.
Let a, b, c be prositive real numbers such that abc # 1, (ab)2 = (bc)4 = (ca)x = abc. Then x equals a) 1
b) 2
c) 3
d) 4
Official Solution :(d)
Explaination from Reso-PCCP side : 70.
given (ab)2 = (bc)4 = (ca)x = abc = k (Let) ab = k1/2 bc = k1/4 ca = k1/x abc = k now. abc = k 2 2 2
(abc)2 = k
1 1 1 . . 2 4 x
= k
k2 2=
1 1 1 . . 2 4 x
1 1 1 . . 2 4 x
1 1 1 . . 2 4 x
x=
4 5
Answer is not in the options PHYSICS : Q. 65. A block of mass 2 Kg placed on a floor experiences an external force in horizontal direction of 20N, frictional force of 6N and norm al force of 20N. The body travels a distance of 10m under the combined effect of all these force. for ce. If Initially body is at rest then what is the kirietic energy of the body at the end of 4m a) 140J
b) 260J
c) 60J
d) 460J
Official Solution :(a)
The work done by Normal force and gravitational force (weight) is zero. Work done by external force is W=20X10=200J. W =20X10=200J. The work done by friction is W / =-6X10=-60J. The net workdone is W=200-60=140J. This workdone results in change in Kinetic energy. 140J=Kf - Ki =Kf-0. Thus Kf = 140J. Explaination from Reso-PCCP side :
By work energy theorem – change in kinetic energy, energy, 'K = work done Kf – Ki = work done done by extranal extranal force + work done by friction Kf – 0 = 20 × 4 – 6 × 4 Kf = 56 J (hence no option is correct) question is bonus is bonus
