DP024 SET 3
Cadangan Jawapan dan Permarkahan
Mock Test (Set 3)
DP024
No Answer Marks
1. (a) (i)
𝜆𝜆 = 𝟐𝟐𝟐𝟐 𝐜𝐜𝐜𝐜 JU1
(ii)
𝑣𝑣 = 𝑓𝑓𝑓𝑓
240 = 𝑓𝑓(20 × 10−2 ) G1
𝑓𝑓 = 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 𝐇𝐇𝐇𝐇 JU1
(iii)
𝑦𝑦 = 𝐴𝐴 sin(𝜔𝜔𝜔𝜔 ± 𝑘𝑘𝑘𝑘)
𝐴𝐴 = 80 cm
= 0.8 m J1
𝜔𝜔 = 2𝜋𝜋𝜋𝜋
= 2𝜋𝜋(1200)
𝜔𝜔 = 2400𝜋𝜋 rad s −1 J1
2𝜋𝜋
𝑘𝑘 =
𝜆𝜆
2𝜋𝜋
𝑘𝑘 =
0.2
𝑘𝑘 = 10 m−1 J1
𝑦𝑦 = 0.8 sin(2400𝜋𝜋𝜋𝜋 − 10𝑥𝑥)
Where 𝑦𝑦 and 𝑥𝑥 are in cm and 𝑡𝑡 is in second JU1
(b) (i)
2𝜋𝜋
𝑘𝑘 =
𝜆𝜆
2𝜋𝜋
62.8 =
𝜆𝜆
𝜆𝜆 = 𝟎𝟎. 𝟏𝟏 𝐦𝐦 GJU1
(ii)
𝜔𝜔 = 2𝜋𝜋𝜋𝜋
314 = 2𝜋𝜋𝜋𝜋
𝑓𝑓 = 𝟒𝟒𝟒𝟒. 𝟗𝟗𝟗𝟗 𝐇𝐇𝐇𝐇 GJU1
DP024 SET 3
(iii)
1
𝑇𝑇 =
𝑓𝑓
1
𝑇𝑇 =
49.79
𝑇𝑇 = 𝟎𝟎. 𝟎𝟎𝟎𝟎 𝐬𝐬 GJU1
(iv)
𝑣𝑣 = 𝑓𝑓𝜆𝜆
𝑣𝑣 = (49.97)(0.1)
𝑣𝑣 = 𝟒𝟒. 𝟗𝟗𝟗𝟗𝟗𝟗 𝐦𝐦 𝐬𝐬 −𝟏𝟏 JU1
Direction: left/ in a negative direction J1
(v)
𝑑𝑑𝑑𝑑
𝑣𝑣 =
𝑑𝑑𝑑𝑑 K1
𝑑𝑑
𝑣𝑣 = [0.001 sin(314𝑡𝑡 + 62.8𝑥𝑥)]
𝑑𝑑𝑑𝑑
R1
𝑣𝑣 = 0.314 cos(314𝑡𝑡 + 62.8𝑥𝑥)
G1
𝑣𝑣 = 0.314 cos[314(3) + 62.8(0.8)]
JU1
𝑣𝑣 = 𝟎𝟎. 𝟐𝟐𝟐𝟐𝟐𝟐 𝐦𝐦 𝐬𝐬−𝟏𝟏
(vi)
𝑣𝑣max = 𝐴𝐴𝜔𝜔
𝑣𝑣max = (0.001)(314)
𝑣𝑣max = 𝟎𝟎. 𝟑𝟑𝟑𝟑𝟑𝟑 𝐦𝐦 𝐬𝐬−𝟏𝟏 GJU1
TOTAL 17
DP024 SET 3
No Answer Marks
2 (a) (i)
𝑑𝑑𝑑𝑑
𝐼𝐼 =
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑 G1
0.4 =
(6.5 × 60)
𝑑𝑑𝑑𝑑 = 𝟏𝟏𝟏𝟏𝟏𝟏 𝐂𝐂 JU1
(ii)
𝑉𝑉 = 𝐼𝐼𝐼𝐼
3 = (0.4)𝑅𝑅 G1
𝑅𝑅 = 𝟕𝟕. 𝟓𝟓 𝛀𝛀 JU1
(b)
𝐴𝐴 = width × thickness K1
= (4 × 10−3 )(9 × 10−6 )
= 3.6 × 10−8 m2
𝜌𝜌𝜌𝜌
𝑅𝑅 =
𝐴𝐴
(3.9 × 10−5 )(10 × 10−3 ) G1
=
3.6 × 10−8 JU1
= 𝟏𝟏𝟏𝟏. 𝟖𝟖𝟖𝟖 𝛀𝛀
(c) (i)
𝑅𝑅2 𝑅𝑅12 = 𝑅𝑅1 + 𝑅𝑅2
= 2+3 G1
= 5Ω
𝑅𝑅3 1 1 1
𝑅𝑅5 𝑅𝑅1 = +
𝑅𝑅123 5 4
𝑅𝑅123 = 2.22 Ω G1
𝑅𝑅4
𝑅𝑅1234 = 5 + 2.22 G1
= 7.22 Ω
1 1 −1
𝑅𝑅𝑒𝑒𝑒𝑒𝑒𝑒 =� + � G1
7.22 1 JU1
= 𝟎𝟎. 𝟖𝟖𝟖𝟖𝟖𝟖 𝛀𝛀
TOTAL 12
DP024 SET 3
No Answer Marks
3. (a) (i)
1 1 1
= +
𝐶𝐶23 𝐶𝐶2 𝐶𝐶3
1 1 −1
𝐶𝐶23 = � + �
22𝜇𝜇 47𝜇𝜇 G1
𝐶𝐶23 = 1.499 × 10−5 F
𝐶𝐶𝑒𝑒𝑒𝑒𝑒𝑒 = 𝐶𝐶1 + 𝐶𝐶23
= 100𝜇𝜇 + 1.499 × 10−5
GJU1
= 𝟏𝟏. 𝟏𝟏𝟏𝟏 × 𝟏𝟏𝟏𝟏−𝟒𝟒 𝐅𝐅
(ii)
𝑉𝑉1 = 𝑉𝑉𝑒𝑒𝑒𝑒𝑒𝑒 = 20 V K1
𝑄𝑄1
𝐶𝐶1 =
𝑉𝑉1
𝑄𝑄1
100𝜇𝜇 =
20 GJU1
𝑄𝑄1 = 𝟐𝟐 × 𝟏𝟏𝟏𝟏−𝟑𝟑 𝐂𝐂
(b) (i)
𝑡𝑡
𝑄𝑄 = 𝑄𝑄𝑜𝑜 𝑒𝑒 −𝑅𝑅𝑅𝑅
𝑄𝑄𝑜𝑜 10
−
2×106 (𝐶𝐶)
G1
= 𝑄𝑄𝑜𝑜 𝑒𝑒
𝑒𝑒 2
𝐶𝐶 = 𝟐𝟐. 𝟓𝟓 × 𝟏𝟏𝟏𝟏−𝟔𝟔 𝐅𝐅 JU1
(ii)
𝜏𝜏 = 𝑅𝑅𝑅𝑅
= (2 × 106 )(2.5 × 10−6 )
= 𝟓𝟓 𝐬𝐬 GJU1
(iii) 𝑡𝑡
𝑄𝑄 = 𝑄𝑄𝑜𝑜 𝑒𝑒 −𝑅𝑅𝑅𝑅
𝑡𝑡
0.5𝑄𝑄𝑜𝑜 = 𝑄𝑄𝑜𝑜 𝑒𝑒 −5 G1
𝑡𝑡 = 𝟑𝟑. 𝟒𝟒𝟒𝟒𝟒𝟒 𝐬𝐬 JU1
TOTAL 9
DP024 SET 3
No Answer Marks
4.(a) 𝜇𝜇0 𝐼𝐼
𝐵𝐵wire =
2𝜋𝜋𝑟𝑟
(4𝜋𝜋 × 10−7 )(8)
𝐵𝐵wire = G1
2𝜋𝜋(8 × 10−2 )
𝐵𝐵wire = 2 × 10−5 T
𝜇𝜇0 𝐼𝐼
𝐵𝐵coil =
2𝑅𝑅
(4𝜋𝜋 × 10−7 )(5) G1
𝐵𝐵coil =
2(12 × 10−2 )
𝐵𝐵coil = 2.618 × 10−5 T
𝐵𝐵net = �(𝐵𝐵wire )2 + (𝐵𝐵coil )2 G1
𝐵𝐵net = �(2 × 10−5 )2 + (2.618 × 10−5 )2
JU1
𝐵𝐵net = 𝟑𝟑. 𝟐𝟐𝟐𝟐𝟐𝟐 × 𝟏𝟏𝟏𝟏−𝟓𝟓 𝐓𝐓
𝐵𝐵coil
𝜃𝜃 = tan−1 � �
𝐵𝐵wire
2.618 × 10−5 G1
= tan−1 � �
2 × 10−5 JU1
𝜃𝜃 = 𝟓𝟓𝟓𝟓. 𝟔𝟔𝟔𝟔°
Above negative z-axis/ Above the horizontal into the page
(b) (i)
𝜀𝜀 = 𝐵𝐵𝐵𝐵𝐵𝐵 sin 𝜃𝜃
𝜀𝜀 = (0.8)(0.2)(12) sin 90° G1
𝜀𝜀 = 1.92 V
𝜀𝜀 = 𝐼𝐼𝐼𝐼
1.92 = 3𝑅𝑅
𝑅𝑅 = 𝟎𝟎. 𝟔𝟔𝟔𝟔 𝛀𝛀 GJU1
(ii)
• 𝐵𝐵ext upwards.
• As the coil moving horizontally to the right, the magnetic flux
in the coil increases. K1
• According to Lenz’s Law 𝐵𝐵Ind will oppose the direction of
𝐵𝐵ext . (𝐵𝐵ind downwards)
• By using RHGR 2, the Induced Current will flow clockwise in
the loop (or from Q to P) J1
DP024 SET 3
(c) (i) Clockwise J1
(ii)
• Area Q is larger that area P.
• The rate of change of magnetic flux in coil Q is higher that coil J1
P.
• Induced 𝜀𝜀 in Q is higher that P.
𝑑𝑑𝜙𝜙 J1
𝜀𝜀 =
𝑑𝑑𝑑𝑑
• Therefore, Q will experience a greater induced current. J1
TOTAL 14
DP024 SET 3
No Answer Marks
5. (a) (i) An image is formed at infinity when 𝑢𝑢 = 𝑓𝑓 K1
𝑅𝑅
𝑓𝑓 =
2
20
=
2
= 𝟏𝟏𝟏𝟏 𝐜𝐜𝐜𝐜 GJU1
(b) (i) K1
𝑀𝑀 = 2
𝑣𝑣
𝑀𝑀 = −
𝑢𝑢
𝑣𝑣
2=−
𝑢𝑢
𝑣𝑣 = −2𝑢𝑢
𝑣𝑣 = −2(1.4)
= −2.8 m G1
1 1 1
= +
𝑓𝑓 𝑢𝑢 𝑣𝑣
1 1 1 G1
= −
𝑓𝑓 1.4 2.8
𝑓𝑓 = 2.8 m
𝑅𝑅 = 2𝑓𝑓
= 2(2.8) GJU1
= 𝟓𝟓. 𝟔𝟔 𝐦𝐦
3
C 𝐹𝐹 𝑂𝑂 𝑃𝑃 𝐼𝐼
1 Front Back
2 rays with arrows
Correct Object Position D3
Correct Image (line
intercept and characteristics
DP024 SET 3
(c) (i)
𝑀𝑀 = −5 K1
𝑣𝑣
𝑀𝑀 = −
𝑢𝑢
𝑣𝑣
−5 = −
𝑢𝑢
𝑣𝑣 = 5𝑢𝑢
|𝑢𝑢| + |𝑣𝑣| = 2.4 m
𝑢𝑢 + 5𝑢𝑢 = 2.4 m G1
𝑢𝑢 = 𝟎𝟎. 𝟒𝟒 𝐦𝐦 JU1
(ii)
1 1 1
= +
𝑓𝑓 𝑢𝑢 𝑣𝑣
1 1 1
= +
𝑓𝑓 𝑢𝑢 5𝑢𝑢
1 1 1
= + G1
𝑓𝑓 0.4 (5)(0.4)
𝑓𝑓 = 𝟎𝟎. 𝟑𝟑𝟑𝟑 𝐦𝐦 JU1
TOTAL 14
DP024 SET 3
No Answer Marks
6. (a) (i)
2𝑦𝑦3 = 20 × 10−2 m
K1
1
�𝑚𝑚 + 2� 𝜆𝜆𝜆𝜆
2� � = 20 × 10−2
𝑑𝑑
1
�3 + 2� 𝜆𝜆(1)
2� � = 20 × 10−2
𝑑𝑑 G1
𝜆𝜆
= 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎 J1
𝑑𝑑
(ii)
1
3𝜆𝜆𝜆𝜆 �1 + 2� 𝜆𝜆𝜆𝜆 K1
𝑦𝑦3 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ𝑡𝑡 − 𝑦𝑦1 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 = −
𝑑𝑑 𝑑𝑑
𝜆𝜆
= (3𝐷𝐷 − 1.5𝐷𝐷)
𝑑𝑑
= (3 − 1.5)(0.0286) G1
= 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎 𝐦𝐦 JU1
(b)
𝑚𝑚 = 19 K1
𝑚𝑚′ = 15 K1
𝑦𝑦19 = 𝑦𝑦15 K1
1
�19 + 2� 𝜆𝜆𝜆𝜆
15𝜆𝜆′𝐷𝐷
=
𝑑𝑑 𝑑𝑑
19.5 (534 × 10−9 ) = 15𝜆𝜆′ G1
𝜆𝜆′ = 𝟔𝟔. 𝟗𝟗𝟗𝟗𝟗𝟗 × 𝟏𝟏𝟏𝟏−𝟕𝟕 𝐦𝐦 JU1
(c)
11∆𝑦𝑦 = 52 × 10−3 m K1
𝜆𝜆𝜆𝜆
11 � � = 52 × 10−3
𝑑𝑑
−9
(633 × 10 )(3)
11 � � = 52 × 10−3 G1
𝑑𝑑
𝑑𝑑 = 4.017 × 10−4 m
𝑑𝑑 = 𝟎𝟎. 𝟒𝟒𝟒𝟒𝟒𝟒 𝐦𝐦𝐦𝐦 JU1
TOTAL 14
